UNKNOWN Questions

Question Number 52950 by gunawan last updated on 15/Jan/19

$$\:\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{3}+\mathrm{sin}\:\mathrm{2}{x}}\:{dx}\:= \\$$

Commented bymaxmathsup by imad last updated on 15/Jan/19

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{sinx}\:+{cosx}}{\mathrm{3}\:+{sin}\left(\mathrm{2}{x}\right)}{dx}\:\Rightarrow\:{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{3}+{sin}\left(\mathrm{2}{x}\right)}{dx} \\$$ $$=_{{x}+\frac{\pi}{\mathrm{4}}={u}} \:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\sqrt{\mathrm{2}}{sin}\left({u}\right)}{\mathrm{3}+{sin}\left(\mathrm{2}\left({u}−\frac{\pi}{\mathrm{4}}\right)\right)}{du}\:=\sqrt{\mathrm{2}}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sinu}}{\mathrm{3}−{cos}\left(\mathrm{2}{u}\right)}{du} \\$$ $$=\sqrt{\mathrm{2}}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sinu}}{\mathrm{3}−\left(\mathrm{2}{cos}^{\mathrm{2}} {u}−\mathrm{1}\right)}{du}\:=\sqrt{\mathrm{2}}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sinu}\:{du}}{\mathrm{4}−\mathrm{2}{cos}^{\mathrm{2}} {u}} \\$$ $$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sinu}}{\mathrm{2}−{cos}^{\mathrm{2}} {u}}{du}\:=_{{cosu}\:={t}} \:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\mathrm{0}} \:\:\frac{−{dt}}{\mathrm{2}−{t}^{\mathrm{2}} }\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{2}−{t}^{\mathrm{2}} } \\$$ $$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}+{t}}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}−{t}}\right){dt}\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\left[{ln}\mid\frac{\sqrt{\mathrm{2}}+{t}}{\sqrt{\mathrm{2}}−{t}}\mid\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \\$$ $$=\frac{\mathrm{1}}{\mathrm{4}}\left\{{ln}\mid\frac{\sqrt{\mathrm{2}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}{\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}\mid\right\}=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{\mathrm{3}}{\mathrm{1}}\right)=\frac{{ln}\left(\mathrm{3}\right)}{\mathrm{4}} \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{d}\left({sinx}−{cosx}\right)}{\mathrm{4}−\mathrm{1}+{sin}\mathrm{2}{x}} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{d}\left({sinx}−{cosx}\right)}{\mathrm{4}−\left({sinx}−{cosx}\right)^{\mathrm{2}} }\:\:\left[{formula}\:\int\frac{{dx}}{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}{a}}{ln}\left(\frac{{a}+{x}}{{a}−{x}}\right)\right] \\$$ $$\frac{\mathrm{1}}{\mathrm{2}×\mathrm{2}}\mid{ln}\left\{\frac{\mathrm{2}+\left({sinx}−{cosx}\right)}{\mathrm{2}−\left({sinx}−{cosx}\right)}\right\}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\$$ $$\frac{\mathrm{1}}{\mathrm{4}}\left[{ln}\left(\frac{\mathrm{2}+\mathrm{0}}{\mathrm{2}−\mathrm{0}}\right)−{ln}\left(\frac{\mathrm{2}+\mathrm{0}−\mathrm{1}}{\mathrm{2}−\mathrm{0}+\mathrm{1}}\right)\right] \\$$ $$=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{0}−{ln}\mathrm{1}+{ln}\mathrm{3}\right] \\$$ $${so}\:{answer}\:{is}\:=\frac{{ln}\mathrm{3}}{\mathrm{4}}\:\:{pls}\:{check}... \\$$

Commented bygunawan last updated on 16/Jan/19

$$\mathrm{Yes}\:\mathrm{Sir},\:\mathrm{nice}\:\mathrm{solution} \\$$

Commented bygunawan last updated on 16/Jan/19

$$\mathrm{Yes}\:\mathrm{Sir},\:\mathrm{nice}\:\mathrm{solution} \\$$

Commented bytanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$${thank}\:{you}... \\$$