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AP EAPCET 16th May 2023 Shift 2 Solved Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

\int\limits_{0}^{\pi} \; \frac{\cos x}{\sqrt{1-\sin^2 x}} dx =

Solution:

Since

\sqrt{1-\sin^2 x}=\sqrt{\cos^2 x}=|\cos x|

, the integrand equals

\frac{\cos x}{|\cos x|}

. Thus it is

1

[0,\pi/2)

and

-1

(\pi/2,\pi]

. The original quotient is undefined at

x=\pi/2

(with finite one-sided limits

1

and

-1

, so it is not a vertical asymptote); split the integral there:

\int_0^\pi \frac{\cos x}{\sqrt{1-\sin^2 x}}\,dx=\int_0^{\pi/2}1\,dx+\int_{\pi/2}^\pi(-1)\,dx=\frac{\pi}{2}-\frac{\pi}{2}=0.

Hence the correct answer is

0

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