Question

What is `int_(pi/8)^((11pi)/12) cos^2xsinx-tan^2xcotx dx`?

Answer 1

`0.60777`

Explanation:

The first step is to separate the integrals.

`int_(pi/8)^((11pi)/12) cos^2xsinxdx - int_(pi/8)^((11pi)/12) tan^2xcotxdx`

`int_(pi/8)^((11pi)/12) cos^2xsinxdx - int_(pi/8)^((11pi)/12) tanx dx`

`int_(pi/8)^((11pi)/12)cos^2xsinxdx - int_(pi/8)^((11pi)/12) sinx/cosx dx`

Let `u = cosx`. Then `du = -sinxdx -> dx= (du)/(-sinx)`. Adjust the bounds of integration accordingly.

`int_(cos(pi/8))^(cos((11pi)/12)) u^2sinx * (du)/(-sinx) - int_(cos(pi/8))^(cos((11pi)/12)) sinx/u * (du)/(-sinx)`

`int_(cos(pi/8))^(cos((11pi)/12)) u^2 du - int_(cos(pi/8))^(cos((11pi)/12)) 1/u du`

These are two trivial integrals.

`[1/3u^3]_(cos(pi/8))^(cos((11pi)/12)) - [ln|u|]_cos(pi/8)^(cos((11pi)/12))`

`[1/3cos^3x]_(cos(pi/8))^(cos((11pi)/12)) - [ln|cosx|]_cos(pi/8)^(cos((11pi)/12)`

This can be evaluated using the second fundamental theorem of calculus as

`0.60777`

Hopefully this helps!