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MHTCET 25 Apr 2025 Shift 2 Paper

Question : 126 of 150

Marks: +1, -0

\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^{x}-1}{e^{x}+1}\right) d x=

Solution:

Concept:

The integrand is an odd function, and the limits of integration are symmetric about zero.

For an odd function, the integral over symmetric limits is zero.

Explanation:

Let

f(x)=\sin \left(\frac{e^{x}-1}{e^{x}+1}\right)

Find

f(-x)

f(-x)=\sin \left(\frac{e^{-x}-1}{e^{-x}+1}\right)

=\sin \left(\frac{1-e^{x}}{1+e^{x}}\right)

=\sin \left[-\left(\frac{e^{x}-1}{e^{x}+1}\right)\right]

=-\sin \left(\frac{e^{x}-1}{e^{x}+1}\right)=-f(x)

Hence,

f(x)

is an odd function.

Also,

\log \frac{1}{2}=-\log 2

So the limits are

-\log 2

and

\log 2

, which are symmetric about zero.

Therefore,

\int_{\log \frac{1}{2}}^{\log 2} f(x)\,dx =\int_{-\log 2}^{\log 2} f(x)\,dx =0

Answer:

0

Option A.

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