Test Index

AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper

Section: Mathematics

© examsnet.com

Question : 32 of 160

Marks: +1, -0

\int\limits_{\pi/6}^{\pi/3} \frac{1}{1+\sqrt{\cot x}}

dx=

$\frac{\pi}{12}$
$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{\pi}{13}$

Solution:

\int\limits_{\pi/6}^{\pi/3} \; \frac{1}{1+\sqrt{\cot x}} d x = I

(say)

I = \int\limits_{\pi/6}^{\pi/3} \; \frac{\sqrt{\cos x}}{\sqrt{\sin x + \sqrt{\cos x}}} d x

...(i)

We have,

\int\limits_{a}^{b} f(x) d x = \int\limits_{0}^{b} f(a+b-x) d x

a+b-x = \; \frac{\pi}{2} - x

I = \int\limits_{\pi/6}^{\pi/3} \; \frac{\sqrt{\cos\left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin\left(\frac{\pi}{2}-x\right)} + \sqrt{\cos\left(\frac{\pi}{2}-x\right)}} d x

I = \int\limits_{\pi/6}^{\pi/3} \; \frac{\sqrt{\sin x}}{\sqrt{\sin x + \sqrt{\cos x}}} d x

...(ii)

Eqs. (i)

+

(ii)

\Rightarrow 2I

= \int\limits_{\pi/6}^{\pi/3} \left( \; \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \; \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) d x

= \int\limits_{\pi/6}^{\pi/3} \left( \; \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) d x

= \int\limits_{\pi/6}^{\pi/3} 1 \cdot d x = (x)_{\pi/6}^{\pi/3} = \left( \; \frac{\pi}{3} - \; \frac{\pi}{6} \right)

2I = \; \frac{\pi}{6} \Rightarrow I = \; \frac{\pi}{12}

© examsnet.com

Go to Question:

Prev Question

Next Question