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Question Number 163288 by HongKing last updated on 05/Jan/22

Answered by Ar Brandon last updated on 05/Jan/22

$$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+a^2{\tan }^{2}x}={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{{\sec }^{2}x+a^2{\tan }^{2}x{\sec }^{2}x}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2+a^2{t}^2(1+t^2)}={\int }_0^{\mathrm{\infty }}\frac{dt}{(t^2+1)(a^2{t}^2+1)}$$$$={\int }_0^{\mathrm{\infty }}(\frac{1}{1-a^2}\cdot \frac{1}{t^2+1}+\frac{a^2}{a^2-1}\cdot \frac{1}{a^2{t}^2+1})dt$$$$=\frac{1}{1-a^2}\cdot \frac{\pi }{2}+\frac{a^2}{a^2-1}\cdot \frac{1}{a}\cdot \frac{\pi }{2}=\frac{1-a}{1-a^2}\cdot \frac{\pi }{2}=\frac{\pi }{2(1+a)}$$

Commented by Ar Brandon last updated on 05/Jan/22

$$\frac{1}{(t^2+1)(a^2{t}^2+1)}=\frac{pt+q}{t^2+1}+\frac{rt+s}{a^2{t}^2+1}$$$$=\frac{(pt+q)(a^2{t}^2+1)+(rt+s)(t^2+1)}{}$$$$\underset{t\to i}{\lim }=(pi+q)(1-a^2)=1,p=0,q=\frac{1}{1-a^2}$$$$q+s=1\Rightarrow s=1-q=\frac{a^2}{a^2-1}$$$$p+r=0\Rightarrow r=0$$

Commented by HongKing last updated on 05/Jan/22

$$\mathrm{perfect}\mathrm{solution}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}$$

Commented by peter frank last updated on 06/Jan/22

$$\mathrm{thank}\mathrm{you}$$

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