Question

Evaluate $\int \frac{sin\theta -cos\theta }{\sqrt{sin2\theta }}d\theta =$

• A. $\frac{-1}{2}ln\left(\frac{sin\theta +cos\theta +\sqrt{sin2\theta }}{sin\theta +cos\theta -\sqrt{sin2\theta }}\right)+c$
• B. $\frac{-1}{4}ln\left(\frac{sin\theta +cos\theta +\sqrt{cos2\theta }}{sin\theta +cos\theta -\sqrt{sin2\theta }}\right)+c$
• C. $\frac{-1}{4}ln\left(\frac{sin\theta -cos\theta +\sqrt{sin2\theta }}{sin\theta +cos\theta -\sqrt{sin2\theta }}\right)+c$
• D. none of these

Answer 1

The correct option is A $\frac{-1}{2}ln\left(\frac{sin\theta +cos\theta +\sqrt{sin2\theta }}{sin\theta +cos\theta -\sqrt{sin2\theta }}\right)+c$

Given, $=I\int \frac{\sin \theta -\cos \theta }{\sqrt{\sin 2\theta }}d\theta$
$=\int \frac{(\sin \theta -\cos \theta )\sec \theta }{\sqrt{2\tan \theta }}d\theta$
$=\int \frac{\tan \theta -1}{\sqrt{2\tan \theta }}d\theta$
Let, $\tan \theta =t^2\Rightarrow {\sec }^2\theta d\theta =2tdt\Rightarrow d\theta (1+t^4)=2tdt$
Substituting back,
$=\int \frac{(t^2-1)2tdt}{\sqrt{2}.t(1+t^4)}$
$=\int \frac{t^2-1}{(1+t^4)}\sqrt{2}dt$
$=\int \frac{t^2-1}{(1+t^4)}\sqrt{2}dt$
$=\int \frac{1-\frac{1}{t^2}}{(\frac{1}{t^2}+t^2)}\sqrt{2}dt$
$=\int \frac{1-\frac{1}{t^2}}{(t+\frac{1}{t}{)}^2-2}\sqrt{2}.dt$
Let, $t+\frac{1}{t}=m=>(1-\frac{1}{t^2})dt=dm$
$=\int \frac{\sqrt{2}dm}{m^2-2}$
$=\int \frac{\sqrt{2}dm}{(m+\sqrt{2})(m-\sqrt{2})}$
$=\int (\frac{-1}{2(m+\sqrt{2})}+\frac{1}{2(m-\sqrt{2})})dm$
$=\frac{-1}{2}ln(m+\sqrt{2})+\frac{1}{2}ln(m-\sqrt{2})+c$
$=\frac{-1}{2}ln\left(\frac{m+\sqrt{2}}{m-\sqrt{2}}\right)+c$
$=\frac{-1}{2}ln\left(\frac{t^2+1+\sqrt{2}t}{t^2+1-\sqrt{2}t}\right)+c$
$=\frac{-1}{2}ln\left(\frac{\tan \theta +1+\sqrt{2tan\theta }}{\tan \theta +1-\sqrt{2\tan \theta }}\right)+c$
$=\frac{-1}{2}ln\left(\frac{\sin \theta +\cos \theta +\sqrt{\sin 2\theta }}{\sin \theta +\cos \theta -\sqrt{\sin 2\theta }}\right)+c$