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GUJCET 2026 Math Paper

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Question : 4 of 40

Marks: +1, -0

\int_0^{\pi} (\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}) dx =

. . . . . . .

$1$
$0$
$-1$
$2$

Solution:

Concept:

Use the identity

\cos 2\theta = \cos^2\theta - \sin^2\theta

to simplify the integrand.

Explanation:

Given integrand is

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

.

This can be written as

-\left(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\right)

.

Using

\cos^2\theta-\sin^2\theta=\cos 2\theta

, with

\theta=\frac{x}{2}

.

So

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

.

Hence the integral becomes

I=\int_0^{\pi} -\cos x\,dx

.

Now

\int -\cos x\,dx=-\sin x

.

Apply limits:

I=[-\sin x]_0^{\pi}

.

So

I=-(\sin\pi-\sin0)

.

Since

\sin\pi=0

and

\sin0=0

, we get

I=-(0-0)=0

.

Answer:

0

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