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Question Number 177820 by cortano1 last updated on 09/Oct/22

$$\underset{\pi /2}{\stackrel{\pi }{\int }}\frac{\mathrm{dx}}{{(\sin \mathrm{x}-2\cos \mathrm{x})}^2}=?$$

Answered by qaz last updated on 09/Oct/22

$${\int }_{\pi /2}^{\pi }\frac{dx}{{(\sin x-2\cos x)}^2}$$$$={\int }_0^{\pi /2}\frac{dx}{{(\cos x+2\sin x)}^2}={\int }_0^{\pi /2}\frac{dx}{\cos {}^{2}x+4\sin x\cos x+4\sin {}^{2}x}$$$$={\int }_0^{\pi /2}\frac{1+\tan {}^{2}x}{1+4\tan x+4\tan {}^{2}x}dx={\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+2x)}^2}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x)}^2}=\frac{1}{2}{\int }_1^{\mathrm{\infty }}\frac{dx}{x^2}=\frac{1}{2x}{\mid }_{\mathrm{\infty }}^1=\frac{1}{2}$$

Answered by Frix last updated on 09/Oct/22

$$\int \frac{dx}{{(\sin x-2\cos x)}^2}\stackrel{t=\frac{1}{2-\tan x}}{=}$$$$\int dt=t=\frac{1}{2-\tan x}+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{1}{2}$$

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