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Question Number 97782 by 175 last updated on 09/Jun/20
Evaluate: ∫ ((sinx)/(1 +sin^2 x))dx
$${Evaluate}: \\ $$$$\int\:\frac{{sinx}}{\mathrm{1}\:+{sin}^{\mathrm{2}} {x}}{dx} \\ $$
Commented by mr W last updated on 09/Jun/20
=−∫((d(cos x))/(((√2)−cos x)((√2)+cos x))) =−((√2)/4)∫[(1/( (√2)−cos x))+(1/( (√2)+cos x))]d(cos x) =((√2)/4)ln (((√2)−cos x)/( (√2)+cos x))+C
$$=−\int\frac{{d}\left(\mathrm{cos}\:{x}\right)}{\left(\sqrt{\mathrm{2}}−\mathrm{cos}\:{x}\right)\left(\sqrt{\mathrm{2}}+\mathrm{cos}\:{x}\right)} \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{cos}\:{x}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{cos}\:{x}}\right]{d}\left(\mathrm{cos}\:{x}\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\frac{\sqrt{\mathrm{2}}−\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{2}}+\mathrm{cos}\:{x}}+{C} \\ $$
Answered by niroj last updated on 09/Jun/20
∫ ((sin x )/(1+ sin^2 x))dx ∫(( sin x dx)/(1+1−cos^2 x)) = ∫ (( sin x dx)/(2−cos^2 x)) Put ,cos x=t −sin x dx= dt sinxdx= −dt −∫ (1/(2−t^2 ))dt = −∫ (( 1)/(((√2) )^2 −(t)^2 ))dt = −(1/(2(√2)))log (( (√2) +t)/( (√2)−t)) +C = −(1/(2(√2))) log (( (√2) +cos x)/( (√2) −cos x)) +C //.
$$\:\:\int\:\frac{\mathrm{sin}\:\mathrm{x}\:}{\mathrm{1}+\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$\:\int\frac{\:\:\mathrm{sin}\:\mathrm{x}\:\mathrm{dx}}{\mathrm{1}+\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:=\:\int\:\frac{\:\mathrm{sin}\:\mathrm{x}\:\mathrm{dx}}{\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\:\mathrm{Put}\:,\mathrm{cos}\:\mathrm{x}=\mathrm{t} \\ $$$$\:\:\:−\mathrm{sin}\:\mathrm{x}\:\mathrm{dx}=\:\mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{sinxdx}=\:−\mathrm{dt} \\ $$$$\:\:−\int\:\frac{\mathrm{1}}{\mathrm{2}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:=\:−\int\:\frac{\:\mathrm{1}}{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} −\left(\mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:=\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{log}\:\frac{\:\sqrt{\mathrm{2}}\:+\mathrm{t}}{\:\sqrt{\mathrm{2}}−\mathrm{t}}\:+\mathrm{C} \\ $$$$\:=\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{log}\:\frac{\:\sqrt{\mathrm{2}}\:\:+\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\mathrm{2}}\:\:−\mathrm{cos}\:\mathrm{x}}\:+\mathrm{C}\://. \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$
Answered by abdomathmax last updated on 09/Jun/20
I =∫ ((sinx dx)/(1+sin^2 x)) ⇒I =∫ ((sinxdx)/(2−cos^2 x)) we do the changement cosx =t ⇒ I =∫ ((−dt)/(2−t^2 )) =∫ (dt/(t^2 −2)) =(1/(2(√2)))∫ ((1/(t−(√2)))−(1/(t+(√2))))dt =(1/(2(√2)))ln∣((t−(√2))/(t+(√2)))∣ +C ⇒ I =(1/(2(√2)))ln∣((cosx−(√2))/(cosx +(√2)))∣ +C
$$\mathrm{I}\:=\int\:\:\frac{\mathrm{sinx}\:\mathrm{dx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:\Rightarrow\mathrm{I}\:=\int\:\:\frac{\mathrm{sinxdx}}{\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{cosx}\:=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int\:\frac{−\mathrm{dt}}{\mathrm{2}−\mathrm{t}^{\mathrm{2}} }\:=\int\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\left(\frac{\mathrm{1}}{\mathrm{t}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{t}+\sqrt{\mathrm{2}}}\right)\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\mathrm{t}−\sqrt{\mathrm{2}}}{\mathrm{t}+\sqrt{\mathrm{2}}}\mid\:+\mathrm{C}\:\:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\mathrm{cosx}−\sqrt{\mathrm{2}}}{\mathrm{cosx}\:+\sqrt{\mathrm{2}}}\mid\:+\mathrm{C} \\ $$

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