Question
If $${u_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^n}\theta \,d\theta } $$ then $${u_n} + {u_{n - 2}}$$ is :
A.
$$\frac{1}{{n - 1}}$$
B.
$$\frac{1}{{n + 1}}$$
C.
$$\frac{1}{{2n - 1}}$$
D.
$$\frac{1}{{2n + 1}}$$
Answer :
$$\frac{1}{{n - 1}}$$
Solution :
$$\eqalign{
& {\text{Given : }}{u_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}\theta \,d\theta } \cr
& \Rightarrow {u_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^2}\theta \,{{\tan }^{n - 2}}\theta \,d\theta } \cr
& \Rightarrow {u_n} = \int\limits_0^{\frac{\pi }{4}} {\left( {{{\sec }^2}\theta - 1} \right)\,{{\tan }^{n - 2}}\theta \,d\theta } \cr
& \Rightarrow {u_n} = \int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}\theta \,{{\tan }^{n - 2}}\theta \,d\theta } - \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}\theta \,d\theta } \cr
& \Rightarrow {u_n} = \int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}\theta \,{{\tan }^{n - 2}}\theta \,d\theta } - {u_{n - 2}} \cr
& \Rightarrow {u_n} + {u_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}\theta \,{{\tan }^{n - 2}}\theta \,d\theta } \cr
& \Rightarrow {u_n} + {u_{n - 2}} = \left. {\frac{{{{\tan }^{n - 1}}\theta }}{{n - 1}}} \right|_0^{\frac{\pi }{4}} \cr
& \Rightarrow {u_n} + {u_{n - 2}} = \frac{1}{{n - 1}} \cr} $$