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Question Number 160767 by cortano last updated on 06/Dec/21

$${\int }_0^{\frac{\pi }{2}}\frac{\cos {}^2\mathrm{x}}{\cos {}^2\mathrm{x}+4\sin {}^2\mathrm{x}}\mathrm{dx}=?$$

Answered by MJS_new last updated on 06/Dec/21

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{{\cos }^{2}x}{{\cos }^{2}x+{4\sin }^{2}x}dx=$$$$[t=\tan x\to dx=\frac{dt}{t^2+1}]$$$$=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{(t^2+1)(4t^2+1)}=\frac{1}{3}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}(\frac{1}{t^2+\frac{1}{4}}-\frac{1}{t^2+1})dt=$$$$=\frac{1}{3}{[2\arctan 2t-\arctan t]}_0^{\mathrm{\infty }}=\frac{\pi }{6}$$

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