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

    ∫_(π/2) ^π  (dx/((sin x−2cos x)^2 )) =?

$$\:\:\:\:\underset{\pi/\mathrm{2}} {\overset{\pi} {\int}}\:\frac{\mathrm{dx}}{\left(\mathrm{sin}\:\mathrm{x}−\mathrm{2cos}\:\mathrm{x}\right)^{\mathrm{2}} }\:=? \\ $$

Answered by qaz last updated on 09/Oct/22

∫_(π/2) ^π (dx/((sin x−2cos x)^2 ))  =∫_0 ^(π/2) (dx/((cos x+2sin x)^2 ))=∫_0 ^(π/2) (dx/(cos^2 x+4sin xcos x+4sin^2 x))  =∫_0 ^(π/2) ((1+tan^2 x)/(1+4tan x+4tan^2 x))dx=∫_0 ^∞ (dx/((1+2x)^2 ))  =(1/2)∫_0 ^∞ (dx/((1+x)^2 ))=(1/2)∫_1 ^∞ (dx/x^2 )=(1/(2x))∣_∞ ^1 =(1/2)

$$\int_{\pi/\mathrm{2}} ^{\pi} \frac{{dx}}{\left(\mathrm{sin}\:{x}−\mathrm{2cos}\:{x}\right)^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dx}}{\left(\mathrm{cos}\:{x}+\mathrm{2sin}\:{x}\right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dx}}{\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{4sin}\:{x}\mathrm{cos}\:{x}+\mathrm{4sin}\:^{\mathrm{2}} {x}} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{4tan}\:{x}+\mathrm{4tan}\:^{\mathrm{2}} {x}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{x}}\mid_{\infty} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Frix last updated on 09/Oct/22

∫(dx/((sin x −2cos x)^2 ))=^(t=(1/(2−tan x)))   ∫dt=t=(1/(2−tan x))+C  ⇒ answer is (1/2)

$$\int\frac{{dx}}{\left(\mathrm{sin}\:{x}\:−\mathrm{2cos}\:{x}\right)^{\mathrm{2}} }\overset{{t}=\frac{\mathrm{1}}{\mathrm{2}−\mathrm{tan}\:{x}}} {=} \\ $$$$\int{dt}={t}=\frac{\mathrm{1}}{\mathrm{2}−\mathrm{tan}\:{x}}+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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