Question

∫sin^5 t/cos^2 t dt

Answer 1

First I split up both the sin^2x into 1-cos^x
So:
I got ∫sinx(1- cos^2x)^2 / cos^2x
Now, I use u substitution;
So I let u = cosx
du = -sinxdx
-du = sinxdx
Now I get:
-∫(1 -u^2)^2/ u^2 du
expanding the top term
-∫(1 -2u^2 + 4u^2) / u^2 du
now simplifying,
-∫u^2 -2 + 1/u^2 du
Now, just integrate:
and I got
-(1/3 u^3 - 2u - 1/u) + c

Answer 2

\[-(u^3/3 - 2u - 1/u) + c\]
\[-((\cos^3x)/3) -2cosx -1/cosx) + c\]

Answer 3

Thx I get it now :D