How do you find the integral of $csc^2x/cot^3x dx$ ?

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Answer 1 · 2018-05-22T09:40:03

$int csc^2x/cot^3x$ $"d"x=1/2tan^2x+"c"$

We want to find $intcsc^2x/cot^3x$ $"d"x$ . Note that

$csc^2x/cot^3x=1/sin^2xsin^3x/cos^3x=sinx/cos^3x=sec^2xtanx$

So,

$intcsc^2x/cot^3x$ $"d"x=intsec^2xtanx$ $"d"x$

We now perform the natural substitution $u=tanx$ and $"d"u=sec^2x$ $"d"x$

Hence,

$intsec^2xtanx$ $"d"x=intu$ $"d"u=1/2u^2=1/2tan^2x+"c"$

How do you find the integral of csc^2x/cot^3x dxcsc^2x/cot^3x dx?