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

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Answer 1 · 2017-02-03T23:05:54

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

Let:

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:

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

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