I am in need of support from the community to help me in figuring out the calculation for the following integral:
$$\int\frac{\cos^5(t)}{(\sin(t))^{\frac{1}{3}}}dt$$
my approach:
$$\cos^5(t) = \cos(t)\left(\frac{1+\cos(2t)}{2}\right)^2 = \frac{1}{4}\bigl[\cos(t)(1+2\cos(2t)+\cos^2(2t))\bigr]$$
Then simplifying the integral:
$$\frac{1}{4}\left[\int \frac{\cos(t)}{(\sin(t))^{\frac{1}{3}}}+2\int\frac{\cos(2t)}{(\sin(t))^{\frac{1}{3}}}+\int\frac{\cos^2(2t)}{(\sin(t))^{\frac{1}{3}}} \right]dt$$
Using $u$ substitution for the first integral as $u = \sin(t)$ and $du = \cos(t)\,dt$
First integral: $\frac{1}{4}\int u^{-\frac{1}{3}}\,du = \frac{1}{4}\left[\frac{3u^{\frac{2}{3}}}{2} \right]$
I'm stuck on the second integral as I cannot make way on how to substitute to get rid of $t$and $2t$ respectively, any ideas?
Best Answer
Substitue $u = \sin t \implies \mathrm{d}u = \cos t \mathrm{d}t$ $$ \int\frac{\cos^5(t)}{(\sin(t))^{\frac{1}{3}}}\mathrm{d}t = \int\frac{\cos^4(t)}{(\sin(t))^{\frac{1}{3}}} \cos t\mathrm{d}t = \int\frac{(1-u^2)^2}{u^{\frac{1}{3}}}\mathrm{d}u = \int (u^{-\frac13} -2u^{\frac53}+u^{\frac{11}{3}})\mathrm{d}u$$