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MHTCET 20 Apr 2026 Shift 2 Paper

Question : 133 of 150

Marks: +1, -0

\int_{-\pi / 2}^{\pi / 2}\left(\sin ^{2} x+\sin ^{3} x\right) d x=k

k

Solution:

Concept:

The integral is simplified by separating the even and odd parts of the integrand.

Explanation:

We have:

k=\int_{-\pi / 2}^{\pi / 2}\left(\sin ^{2} x+\sin ^{3} x\right) d x

Here,

\sin^2 x

is an even function because

\sin^2(-x)=\sin^2 x

And

\sin^3 x

is an odd function because

\sin^3(-x)=-\sin^3 x

For an odd function, the integral over

[-a,a]

is zero, so:

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

Thus, the integral reduces to:

k=\int_{-\pi / 2}^{\pi / 2} \sin^2 x\,dx

Since

\sin^2 x

is even, we write:

k=2\int_{0}^{\pi / 2} \sin^2 x\,dx

Use the identity:

\sin^2 x=\frac{1-\cos 2x}{2}

So:

k=2\int_{0}^{\pi / 2}\frac{1-\cos 2x}{2}\,dx

k=\int_{0}^{\pi / 2}(1-\cos 2x)\,dx

k=\left[x-\frac{\sin 2x}{2}\right]_{0}^{\pi / 2}

k=\frac{\pi}{2}-0=\frac{\pi}{2}

Answer:

k=\frac{\pi}{2}

, which corresponds to option D.

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