Finding the $\int \frac{dx}{\sin^5x}$

Question

$\int \frac{dx}{\sin^5x}$

I did $t= \cos(x)$, $-\frac{dt}{\sin(x)}=dx$:

$$\int \frac{dx}{\sin^5x} = \int \frac{dt}{\sin^6x} = -\int \frac{dt}{(1-t^2)^3}=-\int (1-t^2)^{-3} = \frac{(1-t^2)^{-3}}{-3} = -\frac{1}{3(1-\cos^2x)}+C$$

Is this correct?

Answer 1

If you're familiar with cosecants, you can derive a reduction formula for $$I_n=\int \frac{1}{\sin^{n}(x)}\mathrm{d}x=\int \csc^{n}(x)\mathrm{d}x$$ We only use integration by parts and the identity $1+\cot^2(x)=\csc^2(x)$. $$\begin{align} I_n&=\csc^{n-2}(x)(-\cot(x))-\int (n-2)\csc^{n-3}(x)(-\csc(x)\cot(x))(-\cot(x))\mathrm{d}x\\ &=-\csc^{n-2}(x)\cot(x)-(n-2)\int \csc^{n-2}(x)\cot^2(x)\mathrm{d}x \\ &=-\csc^{n-2}(x)\cot(x)-(n-2)\int \csc^{n-2}(x)(\csc^2(x)-1)\mathrm{d}x \\ &=-\csc^{n-2}(x)\cot(x)-(n-2)\int\csc^{n}(x)\mathrm{d}x+(n-2)\int\csc^{n-2}(x)\mathrm{d}x \end{align}$$ which gives you the formula $$I_n=-\frac{1}{n-1}\csc^{n-2}(x)\cot(x)+\frac{n-2}{n-1}I_{n-2} $$ In your case, apply it for $n=5$ to obtain an expression in terms of $I_3$, and then apply it again for $n=3$. The final step is to use (proved by direct differentiation) $$I_1=\int \csc(x)\mathrm{d}x=\ln|\csc(x)-\cot(x)|+C $$

Answer 2

With $t=\cos x$ \begin{align} \int \frac{1}{\sin^5x} dx= &-\int \frac{1}{(1-t^2)^3}dt =-\int \frac1{4t^3}\ d\left( \frac{t^4}{(1-t^2)^2}\right)\\ \overset{ibp}= & -\frac t{4(1-t^2)^2}-\frac38\int \frac1{t}\ d\left( \frac{t^2}{1-t^2}\right)\\ \overset{ibp}= & - \frac t{4(1-t^2)^2}-\frac38 \left( \frac t{1-t^2}+\tanh^{-1}t \right)+C\\ = & -\frac {\cos x}{4\sin^4 x}- \frac {3\cos x}{8\sin^2x}-\frac38\tanh^{-1}\cos x+C\\ \end{align}