Question
The value of $$\int_0^{\frac{\pi }{2}} {{{\sin }^8}x\,dx} $$ is :
A.
$$\frac{{105\pi }}{{32\left( {4!} \right)}}$$
B.
$$\frac{{105\pi }}{{16\left( {4!} \right)}}$$
C.
$$\frac{{105}}{{16\left( {4!} \right)}}$$
D.
none of these
Answer :
$$\frac{{105\pi }}{{32\left( {4!} \right)}}$$
Solution :
$$\eqalign{
& {I_n} = \int_0^{\frac{\pi }{2}} {{{\sin }^n}x\,dx} \cr
& = \int_0^{\frac{\pi }{2}} {{{\sin }^{n - 1}}x.\sin \,x\,dx} \cr
& = \left[ {{{\sin }^{n - 1}}x.\left( { - \cos \,x} \right)} \right]_0^{\frac{\pi }{2}} - \int_0^{\frac{\pi }{2}} { - \cos \,x.\left( {n - 1} \right){{\sin }^{n - 2}}x.\cos \,x\,dx} \cr
& = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {\left( {{{\sin }^{n - 2}}x - {{\sin }^n}x} \right)dx} \cr
& \therefore \,\,\,n{I_n} = \left( {n - 1} \right){I_{n - 2}} \cr
& \therefore \,\,\,{I_n} = \frac{{n - 1}}{n}{I_{n - 2}} \cr
& \therefore \,\,\,{I_8} = \frac{7}{8}{I_6} = \frac{7}{8}.\frac{5}{6}.\frac{3}{4}.\frac{1}{2}{I_0} = \frac{{1.3.5.7}}{{1.2.3.4}}.\frac{1}{{{2^4}}}.\frac{\pi }{2} \cr} $$