What is the value of the definite integral \int_{0}^{\frac{\pi}{2}} (3\sin x + 2\cos x) \,… — JAMB UTME

What is the value of the definite integral $\int_{0}^{\frac{\pi}{2}} (3\sin x + 2\cos x) \, dx$ ?

The value of the definite integral is

5

The antiderivative of

3\sin x + 2\cos x

-3\cos x + 2\sin x

. Evaluating this expression at the upper limit

x = \frac{\pi}{2}

gives

2

, and at the lower limit

x = 0

gives

-3

. Applying the Fundamental Theorem of Calculus gives

2 - (-3) = 5

Find the indefinite integral of

3\sin x + 2\cos x

\int (3\sin x + 2\cos x) \, dx = -3\cos x + 2\sin x

The antiderivative of

\sin x

-\cos x

and the antiderivative of

\cos x

\sin x

Evaluate the antiderivative at the upper limit

x = \frac{\pi}{2}

-3\cos\left(\frac{\pi}{2}\right) + 2\sin\left(\frac{\pi}{2}\right) = -3(0) + 2(1) = 2

Substitute

x = \frac{\pi}{2}

into the antiderivative expression.

Evaluate the antiderivative at the lower limit

x = 0

-3\cos(0) + 2\sin(0) = -3(1) + 2(0) = -3

Substitute

x = 0

into the antiderivative expression.

Subtract the lower limit value from the upper limit value

2 - (-3) = 2 + 3 = 5

By the Fundamental Theorem of Calculus,

\int_{a}^{b} f(x)\,dx = F(b) - F(a)

Definite Integration of Trigonometric Functions

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