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Q.
If $\int \sin ^{3} x \cos ^{5} x d x=A \sin ^{4} x +B \sin ^{6} x +C \sin ^{8} x+ D$. Then
Integrals
Solution:
$I =\int \sin ^{3} x \cdot \cos ^{5} x dx$
Put $\sin x=t \Rightarrow \cos x d x=d t$
$I =\int \sin ^{3} x \cdot \cos ^{4} x \cdot \cos x d x =\int t ^{3}\left(1- t ^{2}\right)^{2} dt$
$=\int\left(t^{3}-2 t^{5}+t^{7}\right) d t=\frac{1}{4} t^{4}-\frac{2}{6} t^{6}+\frac{1}{8} t^{8}+D$
$=\frac{1}{4} \sin ^{4} x-\frac{1}{3} \sin ^{6} x+\frac{1}{8} \sin ^{8} x+D$