If f(t)= _0^ 2xdx1- ^2t ^2x,0

Question

If $f(t)=\int _{0}^{\pi }\frac{2xdx}{1-{\cos }^{2}t{\sin }^{2}x},0<t<\pi $, then the value of $\int _{0}^{\frac{\pi }{2}}\frac{{\pi }^{2}dt}{f(t)}$ equals \_\_\_\_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

$\begin{gathered} f(t)=\int _{0}^{\pi }\frac{2xdx}{1-{\cos }^{2}t{\sin }^{2}x} \ x	o \pi -x \end{gathered}$

$\begin{gathered} f(t)=\int _{0}^{\pi }\frac{2(\pi -x)dx}{1-{\cos }^{2}t{\sin }^{2}x}=2\pi \int _{0}^{\pi }\frac{dx}{1-{\cos }^{2}t{\sin }^{2}x}-f(t) \ \Rightarrow f(t)=\pi \int _{0}^{\pi }\frac{dx}{1-{\cos }^{2}t{\sin }^{2}x} \ =2\pi \int _{0}^{\frac{\pi }{2}}\frac{dx}{1-{\cos }^{2}t{\sin }^{2}x} \ f(t)=2\pi \int _{0}^{\frac{\pi }{2}}\frac{{\sec }^{2}xdx}{{\sec }^{2}x-{\cos }^{2}t{	an }^{2}x} \ {I}_{1}=\int \frac{{\sec }^{2}xdx}{{\sec }^{2}x-{\cos }^{2}t{	an }^{2}x} \ \cos t	an x=\lambda \Rightarrow \cos t{\sec }^{2}xdx=d\lambda \ {I}_{1}=\int \frac{d\lambda }{\cos t\cdot (1+{\lambda }^{2}{\sec }^{2}t-{\lambda }^{2})}=\int \frac{d\lambda }{\cos t(1+{\lambda }^{2}{	an }^{2}t)} \end{gathered}$

$\begin{gathered} =\frac{1}{\cos t\cdot {	an }^{2}t}\cdot \int \frac{d\lambda }{{\lambda }^{2}+{\cos }^{2}t}= \frac{1}{\cos t{	an }^{2}t} \ 	imes \frac{1}{\cos t}{	an }^{-1}(\lambda 	an t) $

$ =\frac{1}{\sin t}{	an }^{-1}(\sin t	an x) \ \Rightarrow {f(t)=\frac{2\pi }{\sin t}{	an }^{-1}(\sin t	an x)]}_{\frac{\pi }{2}}^{0}=\frac{{\pi }^{2}}{\sin t} \ \Rightarrow \int _{0}^{\frac{\pi }{2}}\frac{{\pi }^{2}dt}{f(t)}=\int _{0}^{\frac{\pi }{2}}\sin tdt=1 \end{gathered}$

Solutions · 1