Question

$\text{If}\int \cot \left(2{\tan }^{-1}\sqrt{\frac{\sqrt{1+\sqrt{x}}-x^{\frac{1}{4}}}{\sqrt{1+\sqrt{x}}+x^{\frac{1}{4}}}}\right)dx=\frac{q.x^{\frac{p}{4}}}{p}+C,x>0$ (where $p$ & $q$ are relatively prime and $C$ is constant of integration), then

• A. $p=5$
• B. $p+2q=13$
• C. $p-q=1$
• D. $q=5$

Answer 1

Answer: (A), (B), (C)

$p-q=1$

Put $x={\tan }^4\theta I=\int \cot \left(2{\tan }^{-1}\sqrt{\frac{\sec \theta -\tan \theta }{\sec \theta +tan\theta }}\right)dxI=\int \cot \left(2{\tan }^{-1}\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}\right)dxI=\int \left(\cot \right(2{\tan }^{-1}\sqrt{\frac{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2}{)}^2}{(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}{)}^2}})dx\int (\cot \left(2{\tan }^{-1}\right(\frac{\tan \frac{\pi }{4}-\tan \frac{\theta }{2}}{1+\tan \frac{\theta }{2}\tan \frac{\pi }{4}}\left)\right))dx=\int \cot (2{\tan }^{-1}\tan (\frac{\pi }{4}-\frac{\theta }{2}))dx=\int \cot (\frac{\pi }{2}-\theta )dx=\int \tan \theta dx=\int x^{\frac{1}{4}}dx=\frac{4x^{\frac{5}{4}}}{5}+C$
On comparing, we get $p=5$ and $q=4$ .