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# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \sin ^{3} x \cos ^{6} x d x$

Solution:

Since power of $\sin$ is odd, put $\cos x=t$

Then $d t=-\sin x d x$

Substitute these in above equation,

$\int \sin ^{3} x \cos ^{6} x d x=\int \sin x \sin ^{2} x t^{6} d x$

$=\int\left(1-\cos ^{2} \mathrm{x}\right) \mathrm{t}^{6} \sin \mathrm{x} \mathrm{d} \mathrm{x}$

$=\int\left(1-\mathrm{t}^{2}\right) \mathrm{t}^{6} \mathrm{dt}$

$=\int\left(\mathrm{t}^{6}-\mathrm{t}^{8}\right) \mathrm{d} \mathrm{t}$

$=\frac{t^{7}}{7}-\frac{t^{9}}{9}+c\left(\right.$ since $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$ for any $\left.c \neq-1\right)$

$=\frac{1}{7} \cos ^{7} x+\frac{1}{9} \cos ^{9} x+c$