Calculate $\int \frac{dx}{1-\sin^4x}$

Question

Calculate $$\int \frac{dx}{1-\sin^4x}$$

My try:

\begin{align} \int \frac{dx}{1-\sin^4x}&=\int \frac{(1-\sin^2x)+(1+\sin^2x)dx}{1-\sin^4x} \\&=\int \frac{dx}{1-\sin^2x}+\int \frac{dx}{1+\sin^2x} \\&=\tan x+\int \frac{dx}{1+\sin^2x} \end{align}

How to deal with the second one?

Answer 1

You appear to be missing a factor of $\frac12$ in front of both integrals because $(1-\sin^2{(x)})+(1+\sin^2{(x)})=2\ne1$ but the integral is question can be found as follows $$\begin{align} \int\frac{\mathrm{d}x}{1+\sin^2{(x)}} &=\int\frac{\csc^2{(x)}}{1+\csc^2{(x)}}\mathrm{d}x\\ &=-\int\frac{\mathrm{d}(\cot{(x)})}{2+\cot^2{(x)}}\mathrm{d}x\\ &=-\frac1{\sqrt{2}}\arctan{\left(\frac{\cot{(x)}}{\sqrt{2}}\right)}+C\\ \end{align}$$

Answer 2

$$\dfrac{dx}{a\sin x\cos x+b\sin^2x+c\cos^2x}$$

Divide numerator and denominator

by $\cos^2x$ and set $\tan x=u$

Or by $\sin^2x$ and set $\cot x=v$

Here $(1)+\sin^2x=(\cos^2x+\sin^2x)+\sin^2x=?$

Answer 3

For the 2nd integral, rewrite the integrand,

$$\frac{1}{1+\sin^2x}=\frac{\sec^2x}{2\sec^2x-1}=\frac{(\tan x)’}{2\tan^2x+1}$$

Then, integrate,

$$\int \frac{d\tan x}{2\tan^2x+1} =\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan x) + C$$