How to integrate this kind of functions? I was wondering how to evaluate $$\int\frac{sin^4 x}{cos^7 x}dx$$
I tried the usual method of writing the expression in terms of powers of $tan(x)$ and $sec(x)$, but nothing useful came out of it.
My attempt
$$\int\frac{sin^4 x}{cos^7 x}dx$$$$=\int({tan^4x}) ({sec^3x})dx$$$$=\int(tan^4x)(sec{x})(sec^2x)dx$$$$=\int(t^4)({\sqrt{t^2+1}})dt$$
I haven't got any further yet.
My generalized question
How to evaluate $$\int(sin^mx)(cos^nx)dx$$
where $$m,n\in \mathbb{Q}$$ and $(m+n)$ is a negative odd integer.
 A: $$\int\frac{\sin^4 x}{\cos^7 x}dx=\int\frac{\sin^4 x\cos{x}}{(1-\sin^2x)^4}dx=$$
$$=\int\frac{\sin^4 x}{(1-\sin^2x)^4}d(\sin{x})$$
A: Hint:
Generalization:
For $I=\int\sin^mx\cos^{2n+1}x\ dx,$
choose $\sin x=u$ to get $$I=\int u^m(1-u^2)^ndu$$
Can you recognize $n$ here?
Similarly for $$\int\cos^mx\sin^{2n+1}x\ dx,$$ choose $\cos x=v$
If the exponent of both $\cos x,\sin x$ are even, use multiple angle formula .
A: \begin{align*}
\int{\frac{sin^{4}x}{cos^{7}x}}\,dx &= \int{tan^{4}x \cdot sec^{3}x}\,dx\\
&= \int{(sec^{2}x - 1)^2 \cdot sec^3{x}}\,dx\\
&= \int{sec^{7}x}\,dx -2\int{sec^{5}x}\,dx + \int{sec^{3}x}\,dx
\end{align*}
Now,
\begin{align*}
I_{2n+1} &= \int{sec^{2n+1}x}\,dx = \int{sec^{2n-1}x\cdot \sec^{2}x}\,dx\\
&= sec^{2n-1}x\int{sec^{2}x}\,dx - \int{((2n-1)sec^{2n-2}x\cdot \sec{x}\cdot\tan{x}\int{sec^{2}x}\,dx})\,dx\\
&= sec^{2n-1}x\cdot \tan{x} - (2n-1)\int{sec^{2n-1}x\cdot tan^{2}x}\,dx\\
&= sec^{2n-1}x\cdot \tan{x} - (2n-1)\int{sec^{2n+1}x}\,dx + (2n-1)\int{sec^{2n-1}x}\,dx\\
&= sec^{2n-1}x\cdot \tan{x} - (2n-1)I_{2n+1} + (2n-1)\int{sec^{2n-1}x}\,dx
\end{align*}
\begin{align*}
&\Rightarrow \quad 2nI_{2n+1} = sec^{2n-1}x\cdot \tan{x} + (2n-1)\int{sec^{2n-1}x}\,dx\\
&\Rightarrow \quad I_{2n+1} = \frac{1}{2n}\left(sec^{2n-1}x\cdot \tan{x} + (2n-1)\int{sec^{2n-1}x}\,dx\right)
\end{align*}
We know, $\int{\sec{x}}\,dx = \log{\vert\sec{x}+\tan{x}\vert} + C$.
Now $n = 1, 2, 3$, give respectively, the expressions of $I_3, I_5, I_7$.
