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TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Section: Mathematics

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Question : 78 of 160

Marks: +1, -0

\int\limits_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1+\cos x}

=

$\pi - 2$
$\pi + 2$
$\frac{\pi}{4}$
$2\sin\frac{\pi}{4}$

Solution:

Let

I=\int\limits_{\;\frac{\pi}{4}}^{\;\frac{3\pi}{4}} \;\frac{dx}{1+\cos x}

. . . (i)

I=\int\limits_{\;\frac{\pi}{4}}^{\;\frac{3\pi}{4}} \;\frac{dx}{1+\cos\left(\;\frac{3\pi}{4}+\;\frac{\pi}{4}-x\right)}=\int\limits_{\;\frac{\pi}{4}}^{\;\frac{3\pi}{4}} \;\frac{dx}{1-\cos x}

. . . (ii)

From Eqs. (i) and (ii),

\Rightarrow 2I=\int\limits_{\;\frac{\pi}{4}}^{\;\frac{3\pi}{4}} \;\frac{1-\cos x+1+\cos x}{(1+\cos x)(1-\cos x)} dx

\Rightarrow 2I=2 \int\limits_{\;\frac{\pi}{4}}^{\;\frac{3\pi}{4}} \;\frac{1}{1-\cos^2 x} dx = \int\limits_{\;\frac{\pi}{4}}^{\;\frac{3\pi}{4}} \csc^2 x \, dx

\Rightarrow I=-[\cot x]_{\;\frac{\pi}{4}}^{\;\frac{3\pi}{4}} = -\left[\cot\left(\;\frac{3\pi}{4}\right)-\cot\left(\;\frac{\pi}{4}\right)\right]

\Rightarrow I=-\left[-\cot \;\frac{\pi}{4} - \cot \;\frac{\pi}{4}\right] = 2 \cot \;\frac{\pi}{4} = 2.1

I=2=2 \sin \;\frac{\pi}{2}

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