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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.

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$$\int\frac{\sin^4 x}{\cos^7 x}dx=\int\frac{\sin^4 x\cos{x}}{(1-\sin^2x)^4}dx=$$ $$=\int\frac{\sin^4 x}{(1-\sin^2x)^4}d(\sin{x})$$

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  • $\begingroup$ Thanks @MichaelRozenberg , and what about the general case? $\endgroup$ – Harry Weasley Oct 15 '17 at 17:22
  • $\begingroup$ @Harry Weasley In the general case it's not so easy, but the starting integral now easy to calculate. $\endgroup$ – Michael Rozenberg Oct 15 '17 at 17:24
  • $\begingroup$ I do agree. I remember learning something about what to do when (m+n) is a negative even number, but I wasn't sure what to do in this case. $\endgroup$ – Harry Weasley Oct 15 '17 at 17:26
  • $\begingroup$ So there's no clear-cut algorithm like there is in the case of (m+n) being a negative even number? $\endgroup$ – Harry Weasley Oct 15 '17 at 17:28
  • $\begingroup$ For given numbers $m$ and $n$ it's possible, but in general case I think it's hard. Maybe I am wrong. $\endgroup$ – Michael Rozenberg Oct 15 '17 at 17:31
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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 .

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\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 $I_3, I_5, I_7$.

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