Question

How do you evaluate the integral of `int (sinx+sinxtan^2x)/sec^2x` to 0 to pi/3?

Answer 1

`1/2`

Explanation:

We have:

`I=int_0^(pi/3)(sinx+sinxtan^2x)/sec^2xdx`

`=int_0^(pi/3)(sinx+sinx(sin^2x/cos^2x))/(1/cos^2x)dx`

`=int_0^(pi/3)cos^2x(sinx+sin^3x/cos^2x)dx`

`=int_0^(pi/3)(cos^2xsinx+sin^3x)dx`

Let `sin^3x=sin^2xsinx=(1-cos^2x)sinx=sinx-cos^2xsinx`:

`I=int_0^(pi/3)(cos^2xsinx+sinx-cos^2xsinx)dx`

`=int_0^(pi/3)sinxdx`

Note that `intsinxdx=-cosx+C`:

`I=[-cosx]_0^(pi/3)`

`=-cos(pi/3)-(-cos(0))`

`=-1/2-(-1)`

`=1/2`