Question

$\int \frac{dx}{1-{\sin }^{4}x}=a\tan x+b{\tan }^{-1}\left(c\tan x\right)+D,$ then

• A. $a=\frac{1}{2}$
• B. $b=\sqrt{2}$
• C. $c=\sqrt{2}$
• D. $b=\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is A $a=\frac{1}{2}$

$\int \frac{1}{1-{\sin }^4\left(x\right)}dx$

$ApplyTrigSubstitution:x=\arctan \left(u\right)$

$=\int \frac{1+u^2}{1+2u^2}du$

$=\int \frac{1}{1+2u^2}du+\int \frac{u^2}{1+2u^2}du$

$=\frac{1}{\sqrt{2}}\arctan \left(\sqrt{2}u\right)-\frac{1}{2\sqrt{2}}\arctan \left(\sqrt{2}u\right)+\frac{1}{2}u$

$=\frac{1}{\sqrt{2}}\arctan \left(\sqrt{2}\tan \left(x\right)\right)-\frac{1}{2\sqrt{2}}\arctan \left(\sqrt{2}\tan \left(x\right)\right)+\frac{1}{2}\tan \left(x\right)$

$=\frac{1}{2}\tan x+\frac{1}{2\sqrt{2}}ta{n}^{-1}\left(\sqrt{2}\tan x\right)+c$

Hence $a=\frac{1}{2}$