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Indefinite Integration of Polynomial and Trigonometric Functions

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Question 1Question →

If ∫(6x2−4sin⁡(2x)) dx=ax3+bcos⁡(2x)+C\int (6x^2 - 4\sin(2x)) \, dx = ax^3 + b\cos(2x) + C, where aa, bb, and CC are constants, what is the value of a+ba + b?

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Answer: 4

Answer

The value of a+ba + b is 44.
Integrating 6x26x^2 gives 2x32x^3, and integrating −4sin⁡(2x)-4\sin(2x) gives 2cos⁡(2x)2\cos(2x). Comparing 2x3+2cos⁡(2x)+C2x^3 + 2\cos(2x) + C to ax3+bcos⁡(2x)+Cax^3 + b\cos(2x) + C yields a=2a=2 and b=2b=2, giving a+b=4a+b=4.

Step-by-Step Solution

1
Integrate each term of the integrand (6x2−4sin⁡(2x))(6x^2 - 4\sin(2x)) with respect to xx.
∫6x2 dx=6x33=2x3\int 6x^2 \, dx = \frac{6x^3}{3} = 2x^3, and ∫−4sin⁡(2x) dx=−4⋅(−12cos⁡(2x))=2cos⁡(2x)\int -4\sin(2x) \, dx = -4 \cdot \left(-\frac{1}{2}\cos(2x)\right) = 2\cos(2x).
Applying the power rule for integration ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1} and the standard trigonometric integral formula ∫sin⁡(kx) dx=−1kcos⁡(kx)\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx).
2
Combine the calculated antiderivative terms and include the constant of integration CC.
∫(6x2−4sin⁡(2x)) dx=2x3+2cos⁡(2x)+C\int (6x^2 - 4\sin(2x)) \, dx = 2x^3 + 2\cos(2x) + C.
Summing the individual term-by-term antiderivatives produces the complete indefinite integral.
3
Compare the resulting expression with ax3+bcos⁡(2x)+Cax^3 + b\cos(2x) + C to determine aa and bb.
a=2a = 2 and b=2b = 2.
Matching corresponding coefficients of x3x^3 and cos⁡(2x)\cos(2x).
4
Compute a+ba + b.
2+2=42 + 2 = 4.
Evaluating the sum of the extracted coefficients.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Estimated Time:1m 30s
Question 2Question →

What is the indefinite integral ∫(4x3−6cos⁡(3x)) dx\int (4x^3 - 6\cos(3x)) \, dx?

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Answer: x4−2sin⁡(3x)+Cx^4 - 2\sin(3x) + C

Answer

x4−2sin⁡(3x)+Cx^4 - 2\sin(3x) + C
Integrating term-by-term, ∫4x3 dx=x4\int 4x^3 \, dx = x^4 and ∫−6cos⁡(3x) dx=−2sin⁡(3x)\int -6\cos(3x) \, dx = -2\sin(3x). Summing these and including the arbitrary constant CC produces x4−2sin⁡(3x)+Cx^4 - 2\sin(3x) + C.

Step-by-Step Solution

1
Integrate the polynomial term 4x34x^3 using the power rule ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
∫4x3 dx=4⋅x44=x4\int 4x^3 \, dx = 4 \cdot \frac{x^4}{4} = x^4
The power rule for integration adds 11 to the exponent and divides by the new exponent.
2
Integrate the trigonometric term −6cos⁡(3x)-6\cos(3x) using ∫cos⁡(kx) dx=1ksin⁡(kx)\int \cos(kx) \, dx = \frac{1}{k}\sin(kx).
∫−6cos⁡(3x) dx=−6⋅13sin⁡(3x)=−2sin⁡(3x)\int -6\cos(3x) \, dx = -6 \cdot \frac{1}{3}\sin(3x) = -2\sin(3x)
Integrating cosine gives positive sine, divided by the coefficient of xx.
3
Combine the results and append the constant of integration CC.
x4−2sin⁡(3x)+Cx^4 - 2\sin(3x) + C
An indefinite integral represents a family of functions and requires the constant of integration CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 3Question →

What is the indefinite integral ∫(6x2−4x2+4sin⁡(2x)cos⁡(2x))dx\int \left( 6x^2 - \frac{4}{x^2} + 4\sin(2x)\cos(2x) \right) dx?

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Answer: 2x3+4x−12cos⁡(4x)+C2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C

Answer

2x3+4x−12cos⁡(4x)+C2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C
Simplifying 4sin⁡(2x)cos⁡(2x)4\sin(2x)\cos(2x) to 2sin⁡(4x)2\sin(4x) using double-angle identities and integrating 6x2−4x−2+2sin⁡(4x)6x^2 - 4x^{-2} + 2\sin(4x) term-by-term yields 2x3+4x−12cos⁡(4x)+C2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C.

Step-by-Step Solution

1
Use the double-angle trigonometric identity to simplify the product term.
4sin⁡(2x)cos⁡(2x)=2(2sin⁡(2x)cos⁡(2x))=2sin⁡(4x)4\sin(2x)\cos(2x) = 2(2\sin(2x)\cos(2x)) = 2\sin(4x)
Applying the double-angle identity sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta transforms the product into a standard single sine term.
2
Express the reciprocal power with a negative exponent.
−4x2=−4x−2-\frac{4}{x^2} = -4x^{-2}
Rewriting the fraction in index form allows the power rule of integration to be applied directly.
3
Integrate each component term using standard integration formulas.
∫6x2dx=2x3\int 6x^2 dx = 2x^3, ∫−4x−2dx=−4x−1−1=4x\int -4x^{-2} dx = \frac{-4x^{-1}}{-1} = \frac{4}{x}, and ∫2sin⁡(4x)dx=2(−14cos⁡(4x))=−12cos⁡(4x)\int 2\sin(4x) dx = 2\left(-\frac{1}{4}\cos(4x)\right) = -\frac{1}{2}\cos(4x)
Integrating xnx^n gives xn+1n+1\frac{x^{n+1}}{n+1} and integrating sin⁡(ax)\sin(ax) yields −1acos⁡(ax)-\frac{1}{a}\cos(ax).
4
Combine the integrated terms and append the constant of integration.
2x3+4x−12cos⁡(4x)+C2x^3 + \frac{4}{x} - \frac{1}{2}\cos(4x) + C
Indefinite integrals require an arbitrary constant CC to represent the entire family of antiderivatives.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 4Question →

What is the indefinite integral ∫(3x2+4cos⁡x) dx\int (3x^2 + 4\cos x) \, dx?

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Answer: x3+4sin⁡x+Cx^3 + 4\sin x + C

Answer

x3+4sin⁡x+Cx^3 + 4\sin x + C
Integrating 3x23x^2 gives x3x^3 via the power rule, and integrating 4cos⁡x4\cos x gives 4sin⁡x4\sin x. Combining these terms along with the required constant of integration CC results in x3+4sin⁡x+Cx^3 + 4\sin x + C.

Step-by-Step Solution

1
Integrate the polynomial term 3x23x^2 using the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}.
∫3x2dx=3x33=x3\int 3x^2 dx = \frac{3x^3}{3} = x^3
The power rule increases the exponent by 1 and divides by the new exponent.
2
Integrate the trigonometric term 4cos⁡x4\cos x using the standard integral ∫cos⁡x dx=sin⁡x\int \cos x \, dx = \sin x.
∫4cos⁡x dx=4sin⁡x\int 4\cos x \, dx = 4\sin x
The antiderivative of cosine is positive sine.
3
Combine the results and append the constant of integration CC.
x3+4sin⁡x+Cx^3 + 4\sin x + C
All indefinite integrals must include an arbitrary constant CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 5Question →

Given that F(x)=∫(3x2−2sin⁡x) dxF(x) = \int (3x^2 - 2\sin x) \, dx and F(0)=6F(0) = 6, what is the value of the constant of integration CC?

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Answer: 4

Answer

The value of the constant of integration CC is 44.
Integrating 3x2−2sin⁡x3x^2 - 2\sin x yields F(x)=x3+2cos⁡x+CF(x) = x^3 + 2\cos x + C. Substituting x=0x = 0 gives F(0)=2(1)+C=2+CF(0) = 2(1) + C = 2 + C. Since F(0)=6F(0) = 6, setting 2+C=62 + C = 6 yields C=4C = 4.

Step-by-Step Solution

1
Integrate the function f(x)=3x2−2sin⁡xf(x) = 3x^2 - 2\sin x with respect to xx
F(x)=x3+2cos⁡x+CF(x) = x^3 + 2\cos x + C
The antiderivative of 3x23x^2 is x3x^3 and the antiderivative of −2sin⁡x-2\sin x is 2cos⁡x2\cos x.
2
Evaluate F(0)F(0) using the antiderivative expression
F(0)=03+2cos⁡(0)+C=2+CF(0) = 0^3 + 2\cos(0) + C = 2 + C
Since cos⁡(0)=1\cos(0) = 1, the term 2cos⁡(0)2\cos(0) simplifies to 22.
3
Solve for the integration constant CC using F(0)=6F(0) = 6
C=4C = 4
Subtracting 22 from both sides of 2+C=62 + C = 6 yields C=4C = 4.

Key Concept

Indefinite integration of polynomial and trigonometric functions with initial conditions
Question 6Question →

What is the indefinite integral ∫((3x+1)2−6sin⁡(3x))dx\int \left( (3x + 1)^2 - 6\sin(3x) \right) dx?

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Answer: 3x3+3x2+x+2cos⁡(3x)+C3x^3 + 3x^2 + x + 2\cos(3x) + C

Answer

3x3+3x2+x+2cos⁡(3x)+C3x^3 + 3x^2 + x + 2\cos(3x) + C
Expanding (3x+1)2(3x+1)^2 yields 9x2+6x+19x^2 + 6x + 1. Integrating 9x2+6x+19x^2 + 6x + 1 gives 3x3+3x2+x3x^3 + 3x^2 + x. Integrating −6sin⁡(3x)-6\sin(3x) gives +2cos⁡(3x)+2\cos(3x). Combining these results with the mandatory constant of integration +C+ C gives 3x3+3x2+x+2cos⁡(3x)+C3x^3 + 3x^2 + x + 2\cos(3x) + C.

Step-by-Step Solution

1
Expand the squared polynomial binomial term
(3x+1)2=9x2+6x+1(3x + 1)^2 = 9x^2 + 6x + 1
Expanding the expression allows for simple term-by-term integration using the power rule.
2
Integrate the polynomial component term by term
∫(9x2+6x+1)dx=3x3+3x2+x\int (9x^2 + 6x + 1) dx = 3x^3 + 3x^2 + x
Apply the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} to each polynomial term.
3
Integrate the trigonometric component
∫−6sin⁡(3x)dx=−6(−13cos⁡(3x))=+2cos⁡(3x)\int -6\sin(3x) dx = -6 \left(-\frac{1}{3}\cos(3x)\right) = +2\cos(3x)
The integral of sin⁡(kx)\sin(kx) is −1kcos⁡(kx)-\frac{1}{k}\cos(kx). Multiplying by −6-6 yields +2cos⁡(3x)+2\cos(3x).
4
Combine terms and add the constant of integration
3x3+3x2+x+2cos⁡(3x)+C3x^3 + 3x^2 + x + 2\cos(3x) + C
Indefinite integrals must include an arbitrary constant +C+ C to account for all antiderivatives.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 7Question →

What is the indefinite integral ∫(3x2−5sin⁡(5x)+2) dx\int (3x^2 - 5\sin(5x) + 2) \, dx?

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Answer: x3+cos⁡(5x)+2x+Cx^3 + \cos(5x) + 2x + C

Answer

x3+cos⁡(5x)+2x+Cx^3 + \cos(5x) + 2x + C
Integrating term-by-term yields ∫3x2dx=x3\int 3x^2 dx = x^3, ∫−5sin⁡(5x)dx=cos⁡(5x)\int -5\sin(5x) dx = \cos(5x), and ∫2dx=2x\int 2 dx = 2x. Adding the arbitrary constant CC produces x3+cos⁡(5x)+2x+Cx^3 + \cos(5x) + 2x + C.

Step-by-Step Solution

1
Integrate the polynomial term 3x23x^2 using the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}
∫3x2dx=3⋅x33=x3\int 3x^2 dx = 3 \cdot \frac{x^3}{3} = x^3
Applying the power rule for integration.
2
Integrate the trigonometric term −5sin⁡(5x)-5\sin(5x) using ∫sin⁡(kx)dx=−1kcos⁡(kx)\int \sin(kx) dx = -\frac{1}{k}\cos(kx)
∫−5sin⁡(5x)dx=−5(−15cos⁡(5x))=cos⁡(5x)\int -5\sin(5x) dx = -5 \left(-\frac{1}{5}\cos(5x)\right) = \cos(5x)
Integration of the sine function reverses differentiation with a positive sign change for negative sine.
3
Integrate the constant term 22 and append the constant of integration CC
∫2dx=2x\int 2 dx = 2x, giving total antiderivative x3+cos⁡(5x)+2x+Cx^3 + \cos(5x) + 2x + C
Indefinite integrals always require an arbitrary constant of integration CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 8Question →

What is the indefinite integral ∫((3−2x)(4x+1)+8cos⁡(4x))dx\int \left( (3 - 2x)(4x + 1) + 8\cos(4x) \right) dx?

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Answer: −83x3+5x2+3x+2sin⁡(4x)+C-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

Answer

−83x3+5x2+3x+2sin⁡(4x)+C-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C
Expanding the product (3−2x)(4x+1)(3 - 2x)(4x + 1) gives −8x2+10x+3-8x^2 + 10x + 3. Integrating term-by-term yields ∫−8x2dx=−83x3\int -8x^2 dx = -\frac{8}{3}x^3, ∫10xdx=5x2\int 10x dx = 5x^2, ∫3dx=3x\int 3 dx = 3x, and ∫8cos⁡(4x)dx=84sin⁡(4x)=2sin⁡(4x)\int 8\cos(4x) dx = \frac{8}{4}\sin(4x) = 2\sin(4x). Adding the integration constant CC gives the complete result −83x3+5x2+3x+2sin⁡(4x)+C-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C.

Step-by-Step Solution

1
Expand the polynomial product inside the integrand
(3−2x)(4x+1)=12x+3−8x2−2x=−8x2+10x+3(3 - 2x)(4x + 1) = 12x + 3 - 8x^2 - 2x = -8x^2 + 10x + 3
Expanding the expression allows term-by-term integration using standard rules.
2
Rewrite the full integrand
∫(−8x2+10x+3+8cos⁡(4x))dx\int \left( -8x^2 + 10x + 3 + 8\cos(4x) \right) dx
Substitute the expanded polynomial back into the integral expression.
3
Integrate each term individually
∫−8x2dx=−83x3\int -8x^2 dx = -\frac{8}{3}x^3, ∫10xdx=5x2\int 10x dx = 5x^2, ∫3dx=3x\int 3 dx = 3x, and ∫8cos⁡(4x)dx=8⋅sin⁡(4x)4=2sin⁡(4x)\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x)
Apply the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} and linear trigonometric rule ∫cos⁡(kx)dx=sin⁡(kx)k\int \cos(kx) dx = \frac{\sin(kx)}{k}.
4
Combine terms and append the constant of integration
−83x3+5x2+3x+2sin⁡(4x)+C-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C
Indefinite integration requires adding an arbitrary constant CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Estimated Time:2m 0s
Question 9Question →

What is the indefinite integral ∫(8x3+12cos⁡(4x)−5) dx\int (8x^3 + 12\cos(4x) - 5) \, dx?

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Answer: 2x4+3sin⁡(4x)−5x+C2x^4 + 3\sin(4x) - 5x + C

Answer

2x4+3sin⁡(4x)−5x+C2x^4 + 3\sin(4x) - 5x + C
Integrating term by term gives ∫8x3 dx=2x4\int 8x^3 \, dx = 2x^4, ∫12cos⁡(4x) dx=3sin⁡(4x)\int 12\cos(4x) \, dx = 3\sin(4x), and ∫−5 dx=−5x\int -5 \, dx = -5x, along with the constant of integration CC, yielding 2x4+3sin⁡(4x)−5x+C2x^4 + 3\sin(4x) - 5x + C.

Step-by-Step Solution

1
Integrate the polynomial term 8x38x^3
8x3+13+1=8x44=2x4\frac{8x^{3+1}}{3+1} = \frac{8x^4}{4} = 2x^4
Apply the power rule for integration: ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
2
Integrate the trigonometric term 12cos⁡(4x)12\cos(4x)
12⋅sin⁡(4x)4=3sin⁡(4x)12 \cdot \frac{\sin(4x)}{4} = 3\sin(4x)
Apply the standard trigonometric integral rule: ∫cos⁡(kx) dx=1ksin⁡(kx)\int \cos(kx) \, dx = \frac{1}{k}\sin(kx).
3
Integrate the constant term −5-5
−5x-5x
The integral of a constant kk with respect to xx is kxkx.
4
Combine all integrated terms and add the arbitrary constant of integration
2x4+3sin⁡(4x)−5x+C2x^4 + 3\sin(4x) - 5x + C
An indefinite integral represents a family of functions and requires the addition of +C+ C.

Key Concept

Indefinite integration of polynomial and trigonometric functions
Estimated Time:1m 30s
Question 10Question →

What is the indefinite integral ∫((2x−3)2+12sin⁡(4x)−6cos⁡(3x))dx\int \left( (2x - 3)^2 + 12\sin(4x) - 6\cos(3x) \right) dx?

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Answer: 43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

Answer

43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C
Expanding (2x−3)2(2x-3)^2 yields 4x2−12x+94x^2 - 12x + 9. Integrating term by term: ∫4x2dx=43x3\int 4x^2 dx = \frac{4}{3}x^3, ∫−12xdx=−6x2\int -12x dx = -6x^2, ∫9dx=9x\int 9 dx = 9x, ∫12sin⁡(4x)dx=−3cos⁡(4x)\int 12\sin(4x) dx = -3\cos(4x), and ∫−6cos⁡(3x)dx=−2sin⁡(3x)\int -6\cos(3x) dx = -2\sin(3x). Summing these results and adding the constant of integration CC gives 43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C.

Step-by-Step Solution

1
Expand the squared binomial inside the integrand.
(2x−3)2=4x2−12x+9(2x - 3)^2 = 4x^2 - 12x + 9
Expanding algebraic terms simplifies the integrand into basic power terms ready for integration.
2
Integrate the polynomial terms using the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}.
\int (4x^2 - 12x + 9) dx = \frac{4}{3}x^3 - 6x^2 + 9x
Applying the power rule to each term individually gives 4x33−12x22+9x\frac{4x^3}{3} - \frac{12x^2}{2} + 9x.
3
Integrate the trigonometric terms using standard integration rules ∫sin⁡(kx)dx=−1kcos⁡(kx)\int \sin(kx) dx = -\frac{1}{k}\cos(kx) and ∫cos⁡(kx)dx=1ksin⁡(kx)\int \cos(kx) dx = \frac{1}{k}\sin(kx).
\int 12\sin(4x) dx = -\frac{12}{4}\cos(4x) = -3\cos(4x) and \int -6\cos(3x) dx = -\frac{6}{3}\sin(3x) = -2\sin(3x)
Integrating sine gives negative cosine divided by the inner linear coefficient, and integrating cosine gives sine divided by the inner linear coefficient.
4
Combine all integrated terms and append the constant of integration CC.
\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C
Indefinite integrals require the addition of an arbitrary constant of integration CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 11Question →

What is the result of evaluating the indefinite integral ∫((2x−1)2−6cos⁡(2x)) dx\int \left( (2x - 1)^2 - 6\cos(2x) \right) \, dx?

Show answer & explanation

Answer: 43x3−2x2+x−3sin⁡(2x)+C\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

Answer

43x3−2x2+x−3sin⁡(2x)+C\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C
Expanding the squared term (2x−1)2(2x - 1)^2 yields 4x2−4x+14x^2 - 4x + 1. Integrating term by term using the power rule gives 43x3−2x2+x\frac{4}{3}x^3 - 2x^2 + x. The integral of −6cos⁡(2x)-6\cos(2x) is −62sin⁡(2x)=−3sin⁡(2x)-\frac{6}{2}\sin(2x) = -3\sin(2x). Adding the arbitrary constant CC results in 43x3−2x2+x−3sin⁡(2x)+C\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C.

Step-by-Step Solution

1
Expand the squared binomial term inside the integral
(2x−1)2=4x2−4x+1(2x - 1)^2 = 4x^2 - 4x + 1
Expanding the expression allows for term-by-term integration using standard integration rules.
2
Integrate each term of the expanded polynomial
\int (4x^2 - 4x + 1) \, dx = \frac{4}{3}x^3 - 2x^2 + x
Apply the power rule for integration: \int x^n \, dx = \frac{x^{n+1}}{n+1}.
3
Integrate the trigonometric term
\int -6\cos(2x) \, dx = -6 \cdot \frac{1}{2}\sin(2x) = -3\sin(2x)
The standard integral of \cos(kx) is \frac{1}{k}\sin(kx).
4
Combine all integrated terms and append the constant of integration
\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C
Indefinite integration requires an arbitrary constant of integration C.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 12Question →

What is the indefinite integral ∫(6x5+8sin⁡(4x)) dx\int (6x^5 + 8\sin(4x)) \, dx?

Show answer & explanation

Answer: x6−2cos⁡(4x)+Cx^6 - 2\cos(4x) + C

Answer

x6−2cos⁡(4x)+Cx^6 - 2\cos(4x) + C
Integrating 6x56x^5 using the power rule gives 6x66=x6\frac{6x^6}{6} = x^6. Integrating 8sin⁡(4x)8\sin(4x) gives 8(−cos⁡(4x)4)=−2cos⁡(4x)8 \left(-\frac{\cos(4x)}{4}\right) = -2\cos(4x). Summing these and appending the constant of integration CC yields x6−2cos⁡(4x)+Cx^6 - 2\cos(4x) + C.

Step-by-Step Solution

1
Split the integral into two individual terms
∫(6x5+8sin⁡(4x)) dx=∫6x5 dx+∫8sin⁡(4x) dx\int (6x^5 + 8\sin(4x)) \, dx = \int 6x^5 \, dx + \int 8\sin(4x) \, dx
Linearity property of integration allows term-by-term integration.
2
Integrate the polynomial term 6x56x^5
\int 6x^5 \, dx = 6 \cdot \frac{x^{5+1}}{5+1} = \frac{6x^6}{6} = x^6
Apply the power rule of integration: ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
3
Integrate the trigonometric term 8sin⁡(4x)8\sin(4x)
\int 8\sin(4x) \, dx = 8 \cdot \left( -\frac{\cos(4x)}{4} \right) = -2\cos(4x)
Apply the standard integral rule: ∫sin⁡(kx) dx=−1kcos⁡(kx)\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx).
4
Combine results and add the constant of integration
x6−2cos⁡(4x)+Cx^6 - 2\cos(4x) + C
Indefinite integration requires an arbitrary constant CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Estimated Time:1m 30s
Question 13Question →

If y=∫(12x2+6sin⁡(2x)) dxy = \int (12x^2 + 6\sin(2x)) \, dx and y=10y = 10 when x=0x = 0, what is the value of the constant of integration CC?

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Answer: 13

Answer

The value of the constant of integration CC is 1313.
Integrating 12x2+6sin⁡(2x)12x^2 + 6\sin(2x) gives y=4x3−3cos⁡(2x)+Cy = 4x^3 - 3\cos(2x) + C. Substituting x=0x = 0 yields y(0)=4(0)3−3cos⁡(0)+C=−3+Cy(0) = 4(0)^3 - 3\cos(0) + C = -3 + C. Setting −3+C=10-3 + C = 10 and solving for CC gives C=13C = 13.

Step-by-Step Solution

1
Integrate the function with respect to xx
y=4x3−3cos⁡(2x)+Cy = 4x^3 - 3\cos(2x) + C
Applying the power rule ∫axn dx=axn+1n+1\int ax^n \, dx = \frac{ax^{n+1}}{n+1} and trigonometric integration rule ∫ksin⁡(bx) dx=−kbcos⁡(bx)\int k\sin(bx) \, dx = -\frac{k}{b}\cos(bx).
2
Apply the initial condition x=0x = 0 and y=10y = 10
10=4(0)3−3cos⁡(0)+C10 = 4(0)^3 - 3\cos(0) + C
Substituting the boundary values to solve for the specific constant of integration.
3
Evaluate trigonometric function and solve for CC
C=13C = 13
Since cos⁡(0)=1\cos(0) = 1, the equation simplifies to 10=−3+C10 = -3 + C, leading directly to C=13C = 13.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Initial Conditions
Question 14Question →

If y=∫(6x−5sin⁡(x)) dxy = \int (6x - 5\sin(x)) \, dx and y=12y = 12 when x=0x = 0, what is the value of the constant of integration CC?

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Answer: 7

Answer

7
Integrating 6x−5sin⁡(x)6x - 5\sin(x) yields 3x2+5cos⁡(x)+C3x^2 + 5\cos(x) + C. Substituting x=0x = 0 and y=12y = 12 gives 12=3(0)2+5(1)+C=5+C12 = 3(0)^2 + 5(1) + C = 5 + C, which gives C=7C = 7.

Step-by-Step Solution

1
Integrate each term of 6x−5sin⁡(x)6x - 5\sin(x) with respect to xx
y=3x2+5cos⁡(x)+Cy = 3x^2 + 5\cos(x) + C
The integral of 6x6x is 3x23x^2, and the integral of −5sin⁡(x)-5\sin(x) is +5cos⁡(x)+5\cos(x).
2
Substitute the initial conditions x=0x = 0 and y=12y = 12 to calculate CC
12=3(0)2+5cos⁡(0)+C  ⟹  12=5+C  ⟹  C=712 = 3(0)^2 + 5\cos(0) + C \implies 12 = 5 + C \implies C = 7
Since cos⁡(0)=1\cos(0) = 1, the right-hand side reduces to 5+C5 + C.

Key Concept

Indefinite Integration and Initial Value Evaluation
Question 15Question →

Find the indefinite integral ∫(3x(x−2)+8cos⁡(4x))dx\int \left( 3x(x - 2) + 8\cos(4x) \right) dx.

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Answer: x3−3x2+2sin⁡(4x)+Cx^3 - 3x^2 + 2\sin(4x) + C

Answer

x3−3x2+2sin⁡(4x)+Cx^3 - 3x^2 + 2\sin(4x) + C
Expanding 3x(x−2)3x(x - 2) yields 3x2−6x3x^2 - 6x. Integrating 3x23x^2 gives x3x^3, integrating −6x-6x yields −3x2-3x^2, and integrating 8cos⁡(4x)8\cos(4x) yields 2sin⁡(4x)2\sin(4x). Adding the arbitrary constant CC produces x3−3x2+2sin⁡(4x)+Cx^3 - 3x^2 + 2\sin(4x) + C.

Step-by-Step Solution

1
Expand the algebraic term inside the integrand.
3x(x−2)=3x2−6x3x(x - 2) = 3x^2 - 6x, expressing the integral as ∫(3x2−6x+8cos⁡(4x))dx\int (3x^2 - 6x + 8\cos(4x)) dx.
Expanding the product converts the term into standard polynomial form suitable for integration.
2
Integrate each component term using basic integration rules.
∫3x2dx=x3\int 3x^2 dx = x^3, ∫−6xdx=−3x2\int -6x dx = -3x^2, and ∫8cos⁡(4x)dx=8⋅sin⁡(4x)4=2sin⁡(4x)\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x).
Apply the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} and trigonometric rule ∫cos⁡(kx)dx=sin⁡(kx)k\int \cos(kx) dx = \frac{\sin(kx)}{k}.
3
Combine terms and attach the constant of integration.
x3−3x2+2sin⁡(4x)+Cx^3 - 3x^2 + 2\sin(4x) + C
An indefinite integral requires an arbitrary constant CC to represent the full set of antiderivatives.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 16Question →

A curve has a gradient function defined by dydx=12x3−6sin⁡(3x)+4\frac{dy}{dx} = 12x^3 - 6\sin(3x) + 4. If the curve passes through the point (0,15)(0, 15), what is the value of the constant of integration CC?

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Answer: 13

Answer

The constant of integration CC is 13.
Integrating 12x3−6sin⁡(3x)+412x^3 - 6\sin(3x) + 4 with respect to xx yields y=3x4+2cos⁡(3x)+4x+Cy = 3x^4 + 2\cos(3x) + 4x + C. Substituting x=0x = 0 and y=15y = 15 gives 15=3(0)4+2cos⁡(0)+4(0)+C=2+C15 = 3(0)^4 + 2\cos(0) + 4(0) + C = 2 + C, which simplifies to C=13C = 13.

Step-by-Step Solution

1
Integrate the gradient function to obtain the general equation of the curve.
y=3x4+2cos⁡(3x)+4x+Cy = 3x^4 + 2\cos(3x) + 4x + C
The integral of 12x312x^3 is 3x43x^4, the integral of −6sin⁡(3x)-6\sin(3x) is +2cos⁡(3x)+2\cos(3x), and the integral of 44 is 4x4x.
2
Apply the initial condition (0,15)(0, 15) to solve for CC.
15=3(0)4+2cos⁡(0)+4(0)+C15 = 3(0)^4 + 2\cos(0) + 4(0) + C
Substitute x=0x = 0 and y=15y = 15 into the integrated equation.
3
Calculate the numerical value of CC.
C=13C = 13
Since cos⁡(0)=1\cos(0) = 1, the equation becomes 15=2+C15 = 2 + C, yielding C=13C = 13.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Initial Conditions
Question 17Question →

If y=∫(4x3−2sin⁡(x)) dxy = \int (4x^3 - 2\sin(x)) \, dx and y=5y = 5 when x=0x = 0, what is the value of the constant of integration CC?

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Answer: 3

Answer

The constant of integration CC is 3.
Integrating 4x3−2sin⁡(x)4x^3 - 2\sin(x) gives y=x4+2cos⁡(x)+Cy = x^4 + 2\cos(x) + C. Substituting x=0x = 0 and y=5y = 5 yields 5=0+2(1)+C5 = 0 + 2(1) + C, so C=3C = 3.

Step-by-Step Solution

1
Integrate the polynomial and trigonometric terms
y=x4+2cos⁡(x)+Cy = x^4 + 2\cos(x) + C
The integral of 4x34x^3 with respect to xx is x4x^4, and the integral of −2sin⁡(x)-2\sin(x) is 2cos⁡(x)2\cos(x).
2
Apply the given initial condition y(0)=5y(0) = 5
5=04+2cos⁡(0)+C  ⟹  5=2+C5 = 0^4 + 2\cos(0) + C \implies 5 = 2 + C
Evaluating the expression at x=0x = 0 uses the trigonometric property cos⁡(0)=1\cos(0) = 1.
3
Solve the linear equation for CC
C=3C = 3
Subtracting 2 from both sides isolates CC.

Key Concept

Determining the Constant of Integration from Initial Boundary Conditions
Question 18Question →

What is the indefinite integral ∫(5x4−3sin⁡(x)) dx\int (5x^4 - 3\sin(x)) \, dx?

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Answer: x5+3cos⁡(x)+Cx^5 + 3\cos(x) + C

Answer

x5+3cos⁡(x)+Cx^5 + 3\cos(x) + C
Integrating 5x45x^4 gives x5x^5, and integrating −3sin⁡(x)-3\sin(x) gives +3cos⁡(x)+3\cos(x) since ∫sin⁡(x) dx=−cos⁡(x)\int \sin(x) \, dx = -\cos(x). Adding the arbitrary constant of integration CC results in x5+3cos⁡(x)+Cx^5 + 3\cos(x) + C.

Step-by-Step Solution

1
Integrate the polynomial term 5x45x^4
∫5x4 dx=5x4+14+1=x5\int 5x^4 \, dx = \frac{5x^{4+1}}{4+1} = x^5
Apply the power rule for integration: ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
2
Integrate the trigonometric term −3sin⁡(x)-3\sin(x)
∫−3sin⁡(x) dx=−3(−cos⁡(x))=3cos⁡(x)\int -3\sin(x) \, dx = -3(-\cos(x)) = 3\cos(x)
The integral of sin⁡(x)\sin(x) with respect to xx is −cos⁡(x)-\cos(x).
3
Combine the terms and add the constant of integration
x5+3cos⁡(x)+Cx^5 + 3\cos(x) + C
An indefinite integral requires an arbitrary constant CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 19Question →

A curve has a gradient function defined by dydx=6x2+8sin⁡(4x)+3\frac{dy}{dx} = 6x^2 + 8\sin(4x) + 3. If the curve passes through the point (0,10)(0, 10), what is the value of the constant of integration CC?

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Answer: 12

Answer

The constant of integration CC is 1212.
Integrating 6x2+8sin⁡(4x)+36x^2 + 8\sin(4x) + 3 with respect to xx gives y=2x3−2cos⁡(4x)+3x+Cy = 2x^3 - 2\cos(4x) + 3x + C. Substituting (0,10)(0, 10) into the integrated equation gives 10=2(0)−2(1)+3(0)+C10 = 2(0) - 2(1) + 3(0) + C, which leads directly to 10=−2+C10 = -2 + C, so C=12C = 12.

Step-by-Step Solution

1
Integrate the gradient function to obtain the general equation of the curve.
y=2x3−2cos⁡(4x)+3x+Cy = 2x^3 - 2\cos(4x) + 3x + C
The antiderivative of sin⁡(kx)\sin(kx) is −1kcos⁡(kx)-\frac{1}{k}\cos(kx) and the antiderivative of xnx^n is xn+1n+1\frac{x^{n+1}}{n+1}.
2
Apply the given initial condition (x,y)=(0,10)(x, y) = (0, 10) to solve for CC.
10=2(0)3−2cos⁡(0)+3(0)+C  ⟹  10=−2+C10 = 2(0)^3 - 2\cos(0) + 3(0) + C \implies 10 = -2 + C
Evaluating at x=0x = 0 requires evaluating cos⁡(0)=1\cos(0) = 1, which leaves −2-2 from the trigonometric term.
3
Solve the linear equation for CC.
C=12C = 12
Adding 22 to both sides of 10=−2+C10 = -2 + C gives C=12C = 12.

Key Concept

Indefinite Integration with Boundary Conditions
Question 20Question →

A curve y=F(x)y = F(x) has a gradient function given by dydx=9x2−8x+6cos⁡(3x)−4sin⁡(2x)\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x). Given that y(0)=7y(0) = 7, determine the value of the constant of integration CC when the antiderivative is expressed in the standard form y=3x3−4x2+2sin⁡(3x)+2cos⁡(2x)+Cy = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C.

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Answer: 5

Answer

The value of the constant of integration CC is 5.
Integrating dydx=9x2−8x+6cos⁡(3x)−4sin⁡(2x)\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x) yields y=3x3−4x2+2sin⁡(3x)+2cos⁡(2x)+Cy = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C. Substituting x=0x = 0 into the expression gives y(0)=0−0+0+2(1)+C=2+Cy(0) = 0 - 0 + 0 + 2(1) + C = 2 + C. Equating to y(0)=7y(0) = 7 gives 2+C=72 + C = 7, which solves to C=5C = 5.

Step-by-Step Solution

1
Integrate the gradient function term-by-term with respect to xx
y=3x3−4x2+2sin⁡(3x)+2cos⁡(2x)+Cy = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C
The antiderivative of 9x29x^2 is 3x33x^3, of −8x-8x is −4x2-4x^2, of 6cos⁡(3x)6\cos(3x) is 2sin⁡(3x)2\sin(3x), and of −4sin⁡(2x)-4\sin(2x) is 2cos⁡(2x)2\cos(2x).
2
Apply the initial boundary condition y(0)=7y(0) = 7
3(0)3−4(0)2+2sin⁡(0)+2cos⁡(0)+C=7  ⟹  2+C=73(0)^3 - 4(0)^2 + 2\sin(0) + 2\cos(0) + C = 7 \implies 2 + C = 7
At x=0x = 0, sin⁡(0)=0\sin(0) = 0 and cos⁡(0)=1\cos(0) = 1, making the non-zero constant contribution equal to 2(1)=22(1) = 2.
3
Solve for the constant of integration CC
C=5C = 5
Subtracting 2 from both sides of 2+C=72 + C = 7 yields C=5C = 5.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Boundary Conditions
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