Question
The following integral $$\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {{{\left( {2\,{\text{cosec}}\,x} \right)}^{17}}dx} $$ is equal to-
A.
$$\int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}du} $$
B.
$$\int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {{{\left( {{e^u} + {e^{ - u}}} \right)}^{17}}du} $$
C.
$$\int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {{{\left( {{e^u} - {e^{ - u}}} \right)}^{17}}du} $$
D.
$$\int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} - {e^{ - u}}} \right)}^{16}}du} $$
Answer :
$$\int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}du} $$
Solution :
$$\eqalign{
& {\text{Let }}I = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {{{\left( {2\,{\text{cosec}}\,x} \right)}^{17}}dx} \cr
& = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {{{\left( {\,{\text{cosec}}\,x + \cot \,x + {\text{cosec}}\,x - \cot \,x} \right)}^{16}}{\text{2}}\,{\text{cosec}}\,x\,dx} \cr
& I = 2\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {{{\left( {{\text{cosec}}\,x + \cot \,x + \frac{1}{{{\text{cosec}}\,x + \cot \,x}}} \right)}^{16}}.\,{\text{cosec}}\,x\,dx} \cr
& {\text{Let cosec}}\,x + \cot \,x = {e^u} \cr
& \Rightarrow \left( { - \,{\text{cosec}}\,x\,\cot \,x - {\text{cosec}}{\,^2}x} \right)dx = {e^u}\,du \cr
& \Rightarrow - \,{\text{cosec}}\,x\,dx = du \cr
& {\text{Also at }}x = \frac{\pi }{4}{\text{,}}\,u = ln\left( {\sqrt 2 + 1} \right) \cr
& {\text{at }}x = \frac{\pi }{2},\,u = ln\,1 = 0 \cr
& \therefore I = - 2\,\int\limits_{ln\left( {\sqrt 2 + 1} \right)}^0 {{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}du} \cr
& = 2\int\limits_0^{ln\left( {\sqrt 2 + 1} \right)} {{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}} du \cr} $$