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JEE Main 26 Feb 2021 Shift 1 Solved Paper

Section: Mathematics

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Question : 79 of 90

Marks: +1, -0

The value of

\int\limits_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^x} \, dx

[26 Feb 2021 Shift 1]

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

Solution:

Let

I=\int\limits_{-\pi/2}^{\pi} \frac{\cos^2 x}{1+3^x} \, dx

\dots

(i)

Using the property,

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

I \; = \int\limits_{-\pi/2}^{\pi/2} \frac{\cos^2(\pi/2 - \pi/2)}{1+3^{\pi/2 - \pi/2}}

\; = \int\limits_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^{-x}} \, dx \;

\; [\because \cos(-x)=\cos x]

I \; = \int\limits_{-\pi/2}^{\pi/2} \frac{3^x \cos^2 x}{1+3^x} \, dx

. . . (ii)

Adding Eqs. (i) and (ii),

2I \; = \int\limits_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^x} \, dx + \int\limits_{-\pi/2}^{\pi/2} \frac{3^x \cos^2 x}{1+3^x} \, dx

\; = \int\limits_{-\pi/2}^{\pi/2} \frac{(1+3^x) \cos^2 x}{1+3^x} \, dx = \int\limits_{-\pi/2}^{\pi/2} \cos^2 x \, dx

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

\; = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_{-\pi/2}^{\pi/2} = \frac{1}{2} [\pi]

\Rightarrow \; 2I \; = \pi/2 \Rightarrow I = \frac{\pi}{4}

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