How do you find the integral of $\frac{1}{{sin}^{2} (x)}$ $1/sin^2(x)$ ?

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Answer 1 · 2016-05-27T18:06:04

Notice that

$dcotx/dx=d(cosx/sinx)/dx=[(cosx)'sinx-cosx*(sinx)']/[sin^2 x]= [-sin^2x-cos^2x]/[sin^2 x]=-1/[(sin^2 x)]$

Hence

$int 1/[sin^2x]dx=-cotx+c$

Answer 2 · 2016-06-04T14:19:18

Maybe more "intuitive" instead of remembering :

$int1/sin^2(x) dx=int (1/cos^2(x))/(sin^2(x)/cos^2(x))dx = int (1/cos^2(x))/tan^2(x)dx$

let's $u = tan(x)$

$du = 1/cos^2(x) dx$

$int 1/u^2du$

which is

$[-1/u]$

and remember that u = tan(x) ::

$[-1/tan(x)]$

How do you find the integral of 1/sin^2(x)1sin2(x) 1/sin^2(x)?