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MHTCET 5 May 2025 Shift 2 Paper

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Question : 124 of 150

Marks: +1, -0

The value of

\int_{0}^{\pi}\left|\sin ^{3} x\right| d x

is

0
$\frac{3}{8}$
$\frac{4}{3}$
$\pi$

Solution:

Concept:

Since

\sin x \ge 0

on

(0,\pi)

, the absolute value sign can be removed.

Explanation:

For

0 < x < \pi

,

\sin x > 0

, hence

|\sin^3 x| = \sin^3 x

.

So the integral becomes

\int_{0}^{\pi} \sin^3 x \, dx

.

Write

\sin^3 x = \sin x \cdot \sin^2 x = \sin x(1 - \cos^2 x)

.

Therefore,

\int_{0}^{\pi} \sin^3 x \, dx = \int_{0}^{\pi} \sin x \, dx - \int_{0}^{\pi} \sin x \cos^2 x \, dx

.

Evaluate each term:

\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = 1 + 1 = 2

.

\int_{0}^{\pi} \sin x \cos^2 x \, dx = \left[-\frac{\cos^3 x}{3}\right]_{0}^{\pi} = \frac{-(-1)^3 + 1^3}{3} = \frac{2}{3}

.

Thus, the integral equals

2 - \frac{2}{3} = \frac{4}{3}

.

Answer:

\frac{4}{3}

Correct option: C.

\frac{4}{3}

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