Question Number 37316 by rahul 19 last updated on 11/Jun/18
$$\int\:\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{dx}\:=\:? \\ $$
Answered by ajfour last updated on 11/Jun/18
![let xsin x+cos x=(√(1+x^2 ))sin (x+tan^(−1) (1/x)) ⇒ (x^2 /((xsin x+cos x)^2 ))=(x^2 /((1+x^2 )sin^2 t)) t=x+tan^(−1) (1/x) dt=[1−(1/x^2 )((1/(1+(1/x^2 ))))]dx ⇒ dt =((x^2 dx)/(1+x^2 )) Hence I=∫cosec^2 tdt =−cot t+c or I=−cot (x+cot^(−1) x)+c .](https://www.tinkutara.com/question/Q37322.png)
$${let}\:{x}\mathrm{sin}\:{x}+\mathrm{cos}\:{x}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mathrm{sin}\:\left({x}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right) \\ $$$$\Rightarrow\:\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\mathrm{sin}\:^{\mathrm{2}} {t}} \\ $$$${t}={x}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}} \\ $$$${dt}=\left[\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\right)\right]{dx} \\ $$$$\Rightarrow\:\:{dt}\:=\frac{{x}^{\mathrm{2}} {dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${Hence}\:{I}=\int\mathrm{cosec}\:^{\mathrm{2}} {tdt} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\mathrm{cot}\:{t}+{c} \\ $$$$\:\:\:\:{or}\:\:{I}=−\mathrm{cot}\:\left({x}+\mathrm{cot}^{−\mathrm{1}} {x}\right)+{c}\:. \\ $$
Commented by rahul 19 last updated on 13/Jun/18
$$\mathrm{Ans}.\:\mathrm{is}\:\frac{−{x}\mathrm{sec}\:{x}}{{x}\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:+\:\mathrm{tan}\:{x}\:+{c}. \\ $$