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Question Number 143962 by lapache last updated on 20/Jun/21

$${\int }_0^{\frac{\pi }{2}}\frac{2{xsin}^{2}t}{{cos}^{2}t+x^2{sin}^{2}t}dt=.....???$$

Answered by Dwaipayan Shikari last updated on 20/Jun/21

$$\frac{2}{x}{\int }_0^{\frac{\pi }{2}}\frac{x^2{sin}^{2}t}{{cos}^{2}t+x^2{sin}^{2}t}dt$$$$=\frac{2}{x}{\int }_0^{\frac{\pi }{2}}1-\frac{{cos}^{2}t}{{cos}^{2}t+x^2{sin}^{2}t}dt=\frac{\pi }{x}-{\int }_0^{\frac{\pi }{2}}\frac{1}{1+x^2{tan}^{2}t}dt$$$$=\frac{\pi }{x}-{\int }_0^{\mathrm{\infty }}\frac{1}{(1+{\left(xu\right)}^2)(1+u^2)}du$$$$=\frac{\pi }{x}+\frac{1}{x^2-1}{\int }_0^{\mathrm{\infty }}\frac{1}{1+x^2{u}^2}-\frac{1}{1+u^2}du$$$$=\frac{\pi }{x}+\frac{1}{x^2-1}(\frac{\pi }{2x}-\frac{\pi }{2})=\frac{\pi }{x}-\frac{\pi }{2x(x+1)}=\frac{\pi }{2x}+\frac{\pi }{(2x+1)}$$

Commented by mnjuly1970 last updated on 20/Jun/21

$$pleaserechechpen-ultimateline$$$$\left(coefficient\right)$$

Answered by ArielVyny last updated on 20/Jun/21

$${\int }_0^{\frac{\pi }{2}}\frac{1}{\frac{{cos}^{2}t+x^2{sin}^{2}t}{2{xsin}^{2}t}}dt$$$${\int }_0^{\frac{\pi }{2}}\frac{1}{\frac{1}{2x}\times \frac{1}{{tan}^{2}t}+\frac{x}{2}}dt=2{\int }_0^{\frac{\pi }{2}}\frac{1}{\frac{1}{{xtan}^{2}t}+x}dt$$$$2{\int }_0^{\frac{\pi }{2}}\frac{1}{\frac{1+{\left(xtant\right)}^2}{{xtan}^{2}t}}=2x{\int }_0^{\frac{\pi }{2}}\frac{{tan}^{2}t}{1+x^2{tan}^{2}t}dt$$$$$$

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