Question

What is `f(x) = int -cos^3x +tanx dx` if `f(pi)=-2`?

Answer 1

`f(x)=-sinx+sin^3x/3-lnabscosx-2`

Explanation:

This can be split into two integrals:

`f(x)=-intcos^3xdx+inttanxdx`

The second integral `inttanxdx=-lnabscosx+C`, this is a common integral.

To find the first, do what follows:

`-intcos^3xdx=-intcos^2xcosxdx`

`=-int(1-sin^2x)cosxdx`

Here, use substitution: let `u=sinx` and `du=cosxdx`. This gives:

`=-int1-u^2du`

Which we can integrate using `intu^ndu=u^(n+1)/(n+1)+C`, going term by term:

`=-u+u^3/3+C=-sinx+sin^3x/3+C`

Thus, the whole expression equals

`f(x)=-sinx+sin^3x/3-lnabscosx+C`

Since `f(pi)=-2`, we see that

`-2=-sinpi+sin^3pi/3-lnabscospi+C`

`-2=-0+0^3/3-lnabs(-1)+C`

`-2=0+0+ln1+C`

`-2=C`

Thus, we obtain

`f(x)=-sinx+sin^3x/3-lnabscosx-2`