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MHTCET 16 May 2024 Shift 1 Paper

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Question : 122 of 150

Marks: +1, -0

The value of

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+2^{x}} d x

is

$\frac{\pi}{4}$
$\frac{\pi}{8}$
$\frac{\pi}{2}$
$4 \pi$

Solution:

Concept:

Use the definite integral property

\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx

and the evenness of

\sin^2 x

.

Explanation:

Let

I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+2^x}\,dx

.

Using the property with

a = -\frac{\pi}{2}

and

b = \frac{\pi}{2}

, we get

a+b=0

.

So,

I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+2^{-x}}\,dx

.

Adding the two expressions for

I

:

2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \left( \frac{1}{1+2^x} + \frac{1}{1+2^{-x}} \right) dx

.

Since

\frac{1}{1+2^{-x}} = \frac{2^x}{1+2^x}

, the bracket equals

1

.

Thus,

2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x\,dx

.

\sin^2 x

is an even function, so

\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx

.

Hence,

2I = 2\int_0^{\frac{\pi}{2}} \sin^2 x\,dx \Rightarrow I = \int_0^{\frac{\pi}{2}} \frac{1-\cos 2x}{2}\,dx

.

I = \frac{1}{2}\left[ x - \frac{\sin 2x}{2} \right]_0^{\frac{\pi}{2}} = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}

.

Answer:

\frac{\pi}{4}

, which is option A.

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