Integration Questions

Question Number 80416 by jagoll last updated on 03/Feb/20

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{x}\mathrm{cos}\:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:{dx}\:? \\$$

Answered by MJS last updated on 03/Feb/20

$$\int\frac{{x}\mathrm{cos}\:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{dx}= \\$$$$\:\:\:\:\:\mathrm{by}\:\mathrm{parts} \\$$$$\:\:\:\:\:{u}={x}\:\rightarrow\:{u}'=\mathrm{1} \\$$$$\:\:\:\:\:{v}'=\frac{\mathrm{cos}\:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:\rightarrow\:{v}=−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:{x}} \\$$$$=−\frac{{x}}{\mathrm{1}+\mathrm{sin}\:{x}}+\int\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}}= \\$$$$=−\frac{{x}}{\mathrm{1}+\mathrm{sin}\:{x}}+\mathrm{tan}\:{x}\:−\frac{\mathrm{1}}{\mathrm{cos}\:{x}}+{C} \\$$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{x}\mathrm{cos}\:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{dx}= \\$$$$\:\:\:\:\:\left[\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\mathrm{tan}\:{x}\:−\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\right)=\mathrm{0}\right] \\$$$$=\mathrm{1}−\frac{\pi}{\mathrm{4}} \\$$

Commented by jagoll last updated on 03/Feb/20

$$\frac{−{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\:\frac{−\frac{\pi}{\mathrm{2}}}{\mathrm{2}}=−\frac{\pi}{\mathrm{4}}. \\$$$${how}\:{get}\:\mathrm{1}−\frac{\pi}{\mathrm{4}}\:{sir}? \\$$

Commented by jagoll last updated on 03/Feb/20

$${oo}\:{i}\:{understand}\:{sir}\:.\:{get}\:\mathrm{1}\: \\$$