Question

The value of ${\int }_0^{\pi /4}\sqrt{\tan \theta }d\theta$ is?

• A. $\frac{\pi }{2\sqrt{2}}(1+\log (3-2\sqrt{2}))$
• B. $\frac{\pi }{2\sqrt{2}}(1+\log (\sqrt{2}-1))$
• C. $\sqrt{\pi }+\log 2$
• D. None of the above

Answer 1

The correct option is D None of the above

Let $I=\int \sqrt{\tan \theta }d\theta =\frac{1}{2}(I_1-I_2)$

Where

$I_1=\int (\tan \theta +\cot \theta )d\theta =\int \frac{\sin \theta +\cos \theta }{\sqrt{\cos \theta \sin \theta }}d\theta$

Put $\sin \theta +\cos \theta =u$

$I_1=\sqrt{2}\int \frac{dt}{\sqrt{{1-t}^2}}=\sqrt{2}{\sin }^{-1}t=\sqrt{2}{\sin }^{-1}(\sin \theta -\cos \theta )$

And $I_2=\int (\sqrt{\cot \theta }-\sqrt{\tan \theta })d\theta$

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

Put $\sin \theta +\cos \theta =t\Rightarrow 2\sin \theta \cos \theta =t^2-1$

$I_2=\sqrt{2}\int \frac{dt}{\sqrt{t^2-1}}=\sqrt{2}\log (1+\sqrt{t^2-1})=\sqrt{2}\log (\sin \theta +\cos \theta +\sqrt{\sin 2\theta })I=\sqrt{2}{\sin }^{-1}(\sin \theta -\cos \theta )+\sqrt{2}\log (\sin \theta +\cos \theta +\sqrt{\sin 2\theta })$

Therefore

${\int }_0^{\pi /4}Id\theta =\sqrt{2}\log (\sqrt{2}+1)-\frac{3\pi }{2}$