What is $int cot^2 (3-x)dx$ ?

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Answer 1 · 2017-01-29T23:05:05

$cot(3-x)-x+C$

$I=intcot^2(3-x)dx$

Use the identity $1+cot^2(theta)=csc^2(theta)$ to say that $cot^2(theta)=csc^2(theta)-1$ .

Then:

$I=intcsc^2(3-x)dx-intdx$

$I=intcsc^2(3-x)dx-x$

For the remaining integral, let $u=3-x$ . This implies that $du=-dx$ . Then:

$I=-intcsc^2(u)du-x$

This is a standard integral, since $d/(du)cot(u)=-csc^2(u)$ :

$I=cot(u)-x$

$I=cot(3-x)-x+C$

What is int cot^2 (3-x)dxint cot^2 (3-x)dx?