$\int \sin ^{4} x \cos ^{3} x d x=\frac{1}{5} \sin ^{5} x-\frac{1}{7} \sin ^{7} x+C$

Question

Examlex · Accepted Answer

The Answer of $\int \sin ^{4} x \cos ^{3} x d x=\frac{1}{5} \sin ^{5} x-\frac{1}{7} \sin ^{7} x+C$

∫sin⁡4xcos⁡3xdx=15sin⁡5x−17sin⁡7x+C\int \sin ^{4} x \cos ^{3} x d x=\frac{1}{5} \sin ^{5} x-\frac{1}{7} \sin ^{7} x+C∫sin4xcos3xdx=51​sin5x−71​sin7x+C