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MHTCET 15 Apr 2026 Shift 1 Paper

Question : 132 of 150

Marks: +1, -0

\int_{-\pi / 2}^{\pi / 2}\left(\frac{x^{2} \cos x}{1+e^{x}}\right) d x

\left(\frac{\pi^{2}}{A}\right)-B

\left(\frac{A}{B}\right)=

Solution:

Concept:

Use the standard property of definite integrals to simplify the integrand with exponential terms.

Explanation:

Let

I=\int_{-\pi / 2}^{\pi / 2}\left(\frac{x^{2} \cos x}{1+e^{x}}\right) d x

Using

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

, we get

I=\int_{-\pi / 2}^{\pi / 2} \frac{x^{2} \cos x}{1+e^{-x}} d x

Adding the two forms of

I

2I=\int_{-\pi / 2}^{\pi / 2} x^{2} \cos x\left(\frac{1}{1+e^{x}}+\frac{1}{1+e^{-x}}\right) d x

Now,

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

Thus,

2I=\int_{-\pi / 2}^{\pi / 2} x^{2} \cos x \, d x

Since

x^{2} \cos x

is an even function,

2I=2\int_{0}^{\pi / 2} x^{2} \cos x \, d x

So,

I=\int_{0}^{\pi / 2} x^{2} \cos x \, d x

Integrating by parts,

\int x^{2} \cos x \, d x = x^{2} \sin x + 2x \cos x - 2 \sin x

Therefore,

I=\left[x^{2} \sin x + 2x \cos x - 2 \sin x\right]_{0}^{\pi / 2}

x=\frac{\pi}{2}

, the expression equals

\frac{\pi^{2}}{4}-2

x=0

, the expression equals

0

Hence,

I=\frac{\pi^{2}}{4}-2

Comparing with

I=\frac{\pi^{2}}{A}-B

A=4, \quad B=2

So,

\frac{A}{B}=\frac{4}{2}=2

Answer:

Option B,

2

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