Question
If $${I_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx} $$ then what is $${I_n} + {I_{n - 2}}$$ equal to ?
A.
$$\frac{1}{n}$$
B.
$$\frac{1}{{\left( {n - 1} \right)}}$$
C.
$$\frac{n}{{\left( {n - 1} \right)}}$$
D.
$$\frac{1}{{\left( {n - 2} \right)}}$$
Answer :
$$\frac{1}{{\left( {n - 1} \right)}}$$
Solution :
$$\eqalign{
& {\text{Let }}{I_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx} \cr
& {\text{Consider,}} \cr
& {I_n} + {I_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx} + \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x\,dx} \cr
& = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x\left( {{{\tan }^2}x + 1} \right)dx} \cr
& = \int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}x\,{{\tan }^{n - 2}}x\,dx} \cr
& {\text{Put }}\tan \,x = t \cr
& {\sec ^2}x\,dx = dt \cr
& {\text{when }}x = 0{\text{ then }}t = 0{\text{ and when }}x = \frac{\pi }{4},\,t = 1 \cr
& \therefore \,{I_n} + {I_{n - 2}} = \int\limits_0^1 {{t^{n - 2}}dt} \cr
& = \left. {\frac{{{t^{n - 2 + 1}}}}{{n - 2 + 1}}} \right|_0^1 \cr
& = \left. {\frac{{{t^{n - 1}}}}{{n - 1}}} \right|_0^1 \cr
& = \frac{1}{{n - 1}}\left[ {1 - 0} \right] \cr
& = \frac{1}{{n - 1}} \cr} $$