How to integrate $\int{\frac{\sin ^{2}\theta }{\cos ^{5}\theta }d\theta }$? How to integrate  $\displaystyle\int\frac{\sin ^{2}\theta }{\cos ^{5}\theta }d\theta$?
This is also  homework,how to start with it?
I try to change into
$$\int\tan ^{2}\theta \sec ^{3}\theta d\theta$$
and then
$$\int\tan ^{}\theta \sec ^{}\theta \; \tan \theta \sec ^{2}\theta d\theta$$
if I set $u=\sec \theta $
then I will deal with a single $ \tan \theta $ then I stuck
 A: $$ \int \dfrac{\sin^2\theta}{\cos^5\theta}\mathscr{d}\theta=\int \dfrac{1}{\cos^5\theta}\mathscr{d}\theta-\int \dfrac{1}{\cos^3\theta}\mathscr{d}\theta$$
Let $I_n:=\int \cos^n\theta\mathscr{d}\theta$
We have
$\begin{align*}
I_n=\int \cos^{n-1}\theta\mathscr{d}\sin\theta&=\cos^{n-1}\theta\sin\theta-\int \sin\theta\mathscr{d}\cos^{n-1}\theta
\\&=\cos^{n-1}\theta\sin\theta+(n-1)\int\cos^{n-2}\theta\sin^2\theta\mathscr{d}\theta
\\&=\cos^{n-1}\theta\sin\theta+(n-1)(I_{n-2}-I_n)
\end{align*}$
Hence, $nI_n=(n-1)I_{n-2}+\cos^{n-1}\theta\sin\theta$.
Also, we have $$I_{-1}=\int \sec\theta\mathscr{d}\theta=\ln |\sec\theta+\tan\theta|+C$$
So,
$$I_{-3}=\dfrac{\ln |\sec\theta+\tan\theta|+\sec^2\theta\sin\theta}{2}+C$$ and $$I_{-5}=\dfrac{3I_{-3}+\sec^4\theta\sin\theta}{4}$$
$$\therefore \int \dfrac{\sin^2\theta}{\cos^5\theta}\mathscr{d}\theta=I_{-5}-I_{-3}=\dfrac{-\ln |\sec\theta+\tan\theta|-\sec^2\theta\sin\theta+2\sec^4\theta\sin\theta}{8}+C$$
A: *

*$\int\sec{x} \ dx=\ln|\sec x+\tan x| , \int \sec^2{x} \ dx=\tan x$ .

*$\int\dfrac{\sin^2 x}{\cos^2 x} \ dx=\int\sec^3 x+\sec^5 x \ dx$ .


If $I=\int \sec^5 x \ dx=\int \sec^3 \sec^2 x \ dx=\int \sec^3 (\tan x)' \ dx$ then $I=\sec^3 x\tan x -\int3\sec^2x \sec x\tan x\tan x \ dx =\\
\sec^3 x\tan x -\int3\sec^3x \tan^2 x=\sec^3 x\tan x -3I+3\int\sec^3x \ dx...$
Do the same for $\int \sec^3 x$ and use 1.
A: for this type of problems, there are some ways of doing it.
you can do this by doing inverse trig identities like setting $\theta$ = $\arcsin(u)$ and don't forget $d\theta$ = $\frac{1}{\sqrt{1-x^2}}$ $du$
then you will get a integral will look like $\int \frac{u^2}{(1-u^2)^3} du$ then you can finish it off but remember your $\theta$ = $\arcsin(u)$ will be now u = $\sin(\theta)$ when you get your answer.
A: It is much simpler to integrate with the substitution $\tan \theta= \sinh t$
$$\int{\frac{\sin ^{2}\theta }{\cos ^{5}\theta }d\theta }
=\frac14\int \sinh^22t \ dt=\frac1{32}\sinh 4t-\frac t8 +C
$$
