Integrate $\int\frac{\sin^2(x)}{\cos^7(x)}dx$

Question

Integrate $\int\frac{\sin^2(x)}{\cos^7(x)}\,dx$

I tried $\int\frac{\sin^2(x)}{\cos^2(x)}\cdot\frac{1}{\cos^5(x)}\,dx$

thean by replacing $\tan^2(x)=t\:\:$ I have $\:\:dx=2\cdot\frac{\sin(x)}{\cos^3(x)}$ but can't get so far with trig manipulations.

Need a bit help if possible :)

Answer 1

hint

Write it as $$\int \frac{\sin^2(x)}{\cos^8(x)}\cos(x)dx$$

$$=\int \frac{\sin^2(x)}{(1-\sin^2(x))^4}d(\sin(x))$$

$$=\int \frac{ t^2dt}{(1-t^2)^4}$$

$$=- \int (\frac{1}{(1-t^2)^3}-\frac{1}{(1-t^2)^4})dt$$

By parts,

$$\int \frac{dt}{(1-t^2)^n}=$$

$$\Bigl[\frac{t}{(1-t^2)^n}\Bigr]+2n\int \frac{1-t^2+1}{(1-t^2)^{n+1}}dt$$

Answer 2

The first idea that I thought of is rewriting the numerator as $1-\cos^2x$; this shows that the integral is equivalent to $$\int\frac{1-\cos^2x}{\cos^7x}dx=\int\sec^7x-\sec^5x~dx$$ Now, these integrals can be solved by using integration by parts, as follows. $$\begin{align}\int\sec^7x~dx&=\int\sec^2x\cdot\sec^5x~dx=\sec^5x\tan x-\int 5\sec^5x\tan x\cdot\tan x~dx\\ &=\sec^5x\tan x-5\int\sec^5x\tan^2x~dx\\ &=\sec^5x\tan x-5\int\sec^5x(\sec^2x-1)~dx\\ &=\sec^5x\tan x-5\int\sec^7x~dx+5\int\sec^5x~dx\end{align}$$ and therefore $$6\int\sec^7x~dx=\sec^5x\tan x+5\int\sec^5x~dx$$ and so on.

This is a pretty tedious method, and to be honest I much prefer hamam-Abdallah's answer.

Answer 3

Rewrite $\int\frac{\sin^2x}{\cos^7x}\,dx = \int \sec^7x dx - \int \sec^5x dx $ and apply the recursive fomula

$$\int \sec^n x \ dx =I_n =\frac{\tan x\sec^{n-2}x}{n-1}+\frac{n-2}{n-1}I_{n-2} $$ to obtain $$\int\frac{\sin^2x}{\cos^7x}\,dx = {\tan x}\left(\frac16\sec^5x -\frac1{24}\sec^3x-\frac1{16}\sec x \right)-\frac1{16}I_0 $$ with $I_0=\int \sec x\ dx = \tanh^{-1}\sin x$.