Question

What is `(int cos^5x dx)/(intsinx dx) -intcos^5x/sinxdx`?

Answer 1

`(sinx-2/3sin^3x+1/5sin^5x+C_1)/(-cosx+C_2)+ln(abssinx)-sin^2x+1/4sin^4x+C_3`

Explanation:

Let:

• `I=intcos^5xdx`
• `J=intsinxdx`
• `K=intcos^5x/sinxdx`

Then:

`I=intcos^4x(cosx)dx=int(cos^2x)^2(cosx)dx`

Rewriting with the Pythagorean identity:

`I=int(1-sin^2x)^2(cosx)dx`

Let `u=sinx` so `du=cosxdx`:

`I=int(1-u^2)^2du=int(1-2u^2+u^4)du`

Term by term:

`I=u-2/3u^3+1/5u^5+C_1=sinx-2/3sin^3x+1/5sin^5x+C_1`

Now:

`J=-cosx+C_2`

And:

`K=intcos^5x/sinxdx`

Using the form of `cos^5x` from above:

`K=int((1-sin^2x)^2cosx)/sinxdx`

Again, `u=sinx` so `du=cosxdx`:

`K=int(1-2u^2+u^4)/udu=int(1/u-2u+u^3)du`

Term by term:

`K=ln(absu)-u^2+1/4u^4=ln(abssinx)-sin^2x+1/4sin^4x+C_3`

Then the original expression is:

`I/J-K=(sinx-2/3sin^3x+1/5sin^5x+C_1)/(-cosx+C_2)+ln(abssinx)-sin^2x+1/4sin^4x+C_3`