question_answer

If \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,dx,}\] prove that \[{{I}_{n}}+{{I}_{n\,+\,2}}=\frac{1}{n+2}.\]            

Answer:

We have, \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,dx,}\]              \[\Rightarrow \] \[{{I}_{n\,+\,2}}=\int_{0}^{\pi /4}{{{\tan }^{n\,+\,2}}x\,dx}\]             \[\Rightarrow \]   \[\,{{I}_{n}}\,+\,{{I}_{n\,+\,2}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,dx}+\int_{0}^{\pi /4}{{{\tan }^{n\,+\,2}}x\,dx}\]             \[=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,(1+{{\tan }^{2}}x)\,dx}\]                                     \[=\int_{0}^{\pi /4}{{{\tan }^{n}}x{{\sec }^{2}}x\,dx}\]             Put       \[\tan \,x=t\] \[\Rightarrow \] \[{{\sec }^{2}}x\,dx=dt\]             Also, when x = 0, then t = 0             And when \[x=\frac{\pi }{4},\] then t = 1             \[\therefore \]      \[{{I}_{n}}+{{I}_{n\,+\,2}}=\int_{0}^{1}{{{t}^{n}}}\,dt=\left[ \frac{{{t}^{n\,+\,1}}}{n+1} \right]_{0}^{1}=\frac{1}{n+1}\] Hence proved.