Answer
$\frac{(cos 2θ)^{\frac{3}{2}}(-7+3(cos 2θ)^{2})}{21}+C$
Work Step by Step
$\int (sin θ)^3 \sqrt{cos 2θ }dθ$
$let 2θ=u$
$2dθ=du$
$dθ=\frac{1}{2}du$
$\int (sin θ)^3 \sqrt{cos 2θ }$
$=\frac{1}{2} \int (cos u)^{\frac{1}{2}}(sin u)^3 du$
$//(sin u)^2= 1-(cos u)^2$
$=\frac{1}{2} \int ((cosu)^{\frac{1}{2}}-(cos u)^{frac{5}{2}}sin u$ $ du$
$=\frac{1}{2} \int sin u(cosu)^{\frac{1}{2}}du$ $-\frac{1}{2} \int (sinu)(cosu)^{\frac{5}{2}}$ $du$
$let cosu=p$
$-sinu$ $du=dp$
$=-\frac{1}{2} \int p^{\frac{1}{2}}dp +\frac{1}{2} \int p^{\frac{5}{2}}dp$
$=-\frac{p^{\frac{3}{2}}}{3}+\frac{p^{\frac{7}{2}}}{7}+C$
$=\frac{-7(cosu)^{\frac{3}{2}}+3(cosu)^{\frac{7}{2}}}{21}+C$
$=\frac{(cosu)^{\frac{3}{2}}(-7+3(cosu)^{2})}{21}+C$
$=\frac{(cos 2θ)^{\frac{3}{2}}(-7+3(cos 2θ)^{2})}{21}+C$
