Test Index

CGPET 2012 Solved Paper

Section: Mathematics

© examsnet.com

Question : 132 of 150

Marks: +1, -0

The value of integral

\int\limits_{0}^{\pi/2} (\sqrt{\tan x}+\sqrt{\cot x}) \, dx

$\sqrt{2} \pi$
$\pi$
$\; \frac{\pi}{2}$
0

Solution:

Let

\;\; I=\int\limits_{0}^{\pi/2} (\sqrt{\tan x}+\sqrt{\cot x}) \, dx

\left[ \begin{array}{l} \because \; \text{ if } \; \int\limits_{0}^{2a} f(x)=\int\limits_{0}^{2a} f(2a-x), \\ \text{ then } \; \int\limits_{0}^{2a} f(x) \, dx=2 \int\limits_{0}^{a} f(x) \, dx \end{array} \right]

\therefore \;\; I=2 \int\limits_{0}^{\pi/4} (\sqrt{\tan x}+\sqrt{\cot x}) \, dx

=2 \int\limits_{0}^{\pi/4} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right) \, dx

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

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

(on multiplying numerator and denominator by

\sqrt{2}

)

=2 \sqrt{2} \int\limits_{0}^{\pi/4} \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}} \, dx

\because (\sin x-\cos x)^2 = \sin^2 x + \cos^2 x.

-2 \sin x \cos x

\therefore \;\; 2 \sin x \cos x = 1 - (\sin x-\cos x)^2 ]

Put

\sin x-\cos x = t \Rightarrow \cos x+\sin x = \frac{dt}{dx}

\Rightarrow \;\; dx = \frac{dt}{\cos x+\sin x}

\therefore \;\; I = 2 \sqrt{2} \int\limits_{-1}^{0} \frac{dt}{\sqrt{1-t^2}}

\; = 2 \sqrt{2} \left[ \sin^{-1} t \right]_{-1}^{0}

= 2 \sqrt{2} \left[ \sin^{-1} 0 - \sin^{-1}(-1) \right]

= 2 \sqrt{2} \left[ 0 + \sin^{-1}(1) \right]

= 2 \sqrt{2} \left( \frac{\pi}{2} \right) = \pi \sqrt{2}

© examsnet.com

Go to Question:

Prev Question

Next Question