_{-π/2}^{π/2}≤ft(x²+ ≤ft(π-x/π+x) · x) d x=

Correct Answer is : π312{π³}{12}12π3

Correct Answer is : π312{π³}{12}12π3

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MHTCET 23 Apr 2025 Shift 2 Paper

Question : 128 of 150

Marks: +1, -0

\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^{2}+\log \left(\frac{\pi-x}{\pi+x}\right) \cdot \cos x\right) d x=

Solution:

Concept:

Use the properties of even and odd functions over a symmetric interval

[-a, a]

Explanation:

Let

f(x) = x^{2} + \log\left(\frac{\pi-x}{\pi+x}\right)\cos x

The term

x^{2}

is an even function.

The term

\log\left(\frac{\pi-x}{\pi+x}\right)

is an odd function, because

\log\left(\frac{\pi+x}{\pi-x}\right) = -\log\left(\frac{\pi-x}{\pi+x}\right)

Since

\cos x

is even, the product

\log\left(\frac{\pi-x}{\pi+x}\right)\cos x

is odd.

The integral of an odd function over

[-\frac{\pi}{2}, \frac{\pi}{2}]

0

Therefore, the given integral reduces to:

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

= 2\left[\frac{x^{3}}{3}\right]_{0}^{\pi/2} = 2 \cdot \frac{1}{3} \cdot \frac{\pi^{3}}{8} = \frac{\pi^{3}}{12}

Answer:

\frac{\pi^{3}}{12}

, which is option B.

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