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AIEEE 2005 Math Solved Paper

Section: Mathematics

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Question : 42 of 71

Marks: +1, -0

The value of

\int\limits_{-\pi}^{\pi} \frac{\cos ^2}{1+a^x}\,dx, a>0

$a\pi$
$\frac{\pi}{2}$
$\frac{\pi}{a}$
$2\pi$

Solution:

Let

I=\int\limits_{-\pi}^{\pi} \frac{\cos ^2 x}{1+a^x}\,dx

......(1)

=\int\limits_{-\pi}^{\pi} \frac{\cos ^2(-x)}{\overset{-x}{1+a}}\,dx

[ Using

\int\limits_{a}^{b} f(x)\,dx=\int\limits_{a}^{b} f(a+b-x)\,dx

]

=\int\limits_{-\pi}^{\pi} \frac{\cos ^2 x}{1+\alpha^2}\,dx

.....(2)

Adding equations (1) and (2) we get

2I=\int\limits_{-\pi}^{\pi} \cos^2 x \left(\frac{1+a^x}{1+a^x}\right)\,dx

=\int\limits_{-\pi}^{\pi} \cos^2 x\,dx

=2\int\limits_{0}^{\pi} \cos^2 x\,dx

=2 \times 2 \int\limits_{0}^{\frac{\pi}{2}} \cos^2 x\,dx

=4 \int\limits_{0}^{\frac{\pi}{2}} \sin^2 x\,dx

\Rightarrow I=2 \int\limits_{0}^{\frac{\pi}{2}} \sin^2 x\,dx

=2 \int\limits_{0}^{\frac{\pi}{2}} (1-\cos^2 x)\,dx

\Rightarrow I=2 \int\limits_{0}^{\frac{\pi}{2}} dx - 2 \int\limits_{0}^{\frac{\pi}{2}} \cos^2 x\,dx

\Rightarrow I+I=2 \left(\frac{\pi}{2}\right)=\pi

\Rightarrow I=\frac{\pi}{2}

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