Integration

**ff4930** · October 4th 2007, 02:21 PM

Hello everyone,

Im having difficulties with this integral.

Integral x csc x cot x dx

It cant be subsitution because dy/dx of csc x is -cscx and cotx.
It got to be integration by parts but i have x csc x and cot x,
anyone can give me a little bit so I can start it off. Thanks.

**Krizalid** · October 4th 2007, 02:49 PM

Originally Posted by ff4930

Integral x csc x cot x dx

$\int x\csc x\cot x\,dx=\int\frac{x\cos x}{\sin^2x}\,dx$

Define $\alpha=\sin x$

Next step is integration by parts.

**ff4930** · October 4th 2007, 03:22 PM

ugh, Im still stumped. I know how you got to where you are tho.

Can I move the sin^2x up and become (sin^2x)^-1?
let u be xcosx and dv = (sin^2x)^-1

**Krizalid** · October 4th 2007, 05:28 PM

$\int\frac{x\cos x}{\sin^2x}\,dx$

$\alpha=\sin x\implies d\alpha=\cos x\,dx,$ the integral becomes to

$\int\frac{x\cos x}{\sin^2x}\,dx=\int\frac{\arcsin\alpha}{\alpha^2} \,d\alpha$

Now let's apply integration by parts, so set $u=\arcsin\alpha\implies du=\frac1{\sqrt{1-\alpha^2}}\,d\alpha$ and $dv=\alpha^{-2}\,d\alpha\implies v=-\frac1\alpha,$ the new integral becomes to

$\int\frac{\arcsin\alpha}{\alpha^2}\,d\alpha=-\frac{\arcsin\alpha}\alpha+\int\frac1{\alpha\sqrt{ 1-\alpha^2}}\,d\alpha$

The remaining integral is easy to solve, so its answer is

$\ln|\alpha|-\ln\left(\sqrt{1-\alpha^2}+1\right)$

Finally

$\int\frac{\arcsin\alpha}{\alpha^2}\,d\alpha=-\frac{\arcsin\alpha}\alpha+\ln|\alpha|-\ln\left(\sqrt{1-\alpha^2}+1\right)$

Back substitute and you are done.

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