Question Number 50414 by Abdo msup. last updated on 16/Dec/18
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{x}\:{sinx}\:{cosx}}{{tan}^{\mathrm{2}} {x}\:+{cotan}^{\mathrm{2}} {x}}{dx} \\ $$$${ctanx}\:=\frac{\mathrm{1}}{{tanx}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18
![I=∫_0 ^(π/2) ((((π/2)−x)cosxsinx)/(cot^2 x+tan^2 x))dx 2I=∫_0 ^(π/2) ((((π/2)−x)cosxsinx+xsinxcosx)/(cot^2 x+tan^2 x))dx 2I=∫_0 ^(π/2) (((π/2)sinxcosx)/(tan^2 x+cot^2 x))dx 2I=(π/2)∫_0 ^(π/2) ((tanxcos^2 x)/(tan^2 x+(1/(tan^2 x))))dx now I_1 =∫((tanx)/((1+tan^2 x)(tan^2 x+(1/(tan^2 x)))))dx t=tanx dt=sec^2 x dx (dt/(1+t^2 ))=dx ∫(t/((1+t^2 )(t^2 +(1/t^2 ))))×(dt/(1+t^2 )) ∫(t^3 /((1+t^2 )^2 (t^4 +1)))dt k=t^2 dk=2tdt (dk/2)=tdt ∫((k×(dk/2))/((1+k)^2 (1+k^2 ))) (1/4)∫((2kdk)/((1+k)^2 (1+k^2 ))) (1/4)∫(((1+k)^2 −(1+k^2 ))/((1+k)^2 (1+k^2 )))dk (1/4)∫(dk/(1+k^2 ))−(1/4)∫(dk/((1+k)^2 )) =(1/4)tan^(−1) (k)−(1/4)×(((1+k)^(−2+1) )/(−2+1)) =(1/4)tan^(−1) (k)+(1/(4(1+k))) =(1/4)tan^(−1) (t^2 )+(1/(4(1+t^2 ))) =(1/4)tan^(−1) (tan^2 x)+(1/(4(1+tan^2 x))) 2I=(π/2)×∣(1/4)tan^(−1) (tan^2 x)+(1/(4(1+tan^2 x)))∣_0 ^(π/2) 2I=(π/8)[{(tan^(−1) (tan^2 (π/2))+(1/(1+tan^2 (π/2))))}−{(tan^(−1) (0)+(1/(1+0)))}] I=(π/(16))[(tan^(−1) (∞)+(1/(1+∞)))−(1)] I=(π/(16))((π/2)+0−1) I=(π^2 /(32))−(π/(16)) pls check...](https://www.tinkutara.com/question/Q50545.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\frac{\pi}{\mathrm{2}}−{x}\right){cosxsinx}}{{cot}^{\mathrm{2}} {x}+{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\frac{\pi}{\mathrm{2}}−{x}\right){cosxsinx}+{xsinxcosx}}{{cot}^{\mathrm{2}} {x}+{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\frac{\pi}{\mathrm{2}}{sinxcosx}}{{tan}^{\mathrm{2}} {x}+{cot}^{\mathrm{2}} {x}}{dx} \\ $$$$\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tanxcos}^{\mathrm{2}} {x}}{{tan}^{\mathrm{2}} {x}+\frac{\mathrm{1}}{{tan}^{\mathrm{2}} {x}}}{dx} \\ $$$${now}\:{I}_{\mathrm{1}} =\int\frac{{tanx}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\left({tan}^{\mathrm{2}} {x}+\frac{\mathrm{1}}{{tan}^{\mathrm{2}} {x}}\right)}{dx} \\ $$$${t}={tanx}\:\:{dt}={sec}^{\mathrm{2}} {x}\:{dx} \\ $$$$\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }={dx} \\ $$$$\int\frac{{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}×\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\int\frac{{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left({t}^{\mathrm{4}} +\mathrm{1}\right)}{dt} \\ $$$${k}={t}^{\mathrm{2}} \:\:\:{dk}=\mathrm{2}{tdt}\:\:\:\:\:\frac{{dk}}{\mathrm{2}}={tdt} \\ $$$$\int\frac{{k}×\frac{{dk}}{\mathrm{2}}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} \left(\mathrm{1}+{k}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{2}{kdk}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} \left(\mathrm{1}+{k}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} −\left(\mathrm{1}+{k}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} \left(\mathrm{1}+{k}^{\mathrm{2}} \right)}{dk} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dk}}{\mathrm{1}+{k}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dk}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{tan}^{−\mathrm{1}} \left({k}\right)−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\left(\mathrm{1}+{k}\right)^{−\mathrm{2}+\mathrm{1}} }{−\mathrm{2}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{tan}^{−\mathrm{1}} \left({k}\right)+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{k}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{tan}^{−\mathrm{1}} \left({t}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{tan}^{−\mathrm{1}} \left({tan}^{\mathrm{2}} {x}\right)+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)} \\ $$$$\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}×\mid\frac{\mathrm{1}}{\mathrm{4}}{tan}^{−\mathrm{1}} \left({tan}^{\mathrm{2}} {x}\right)+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{2}{I}=\frac{\pi}{\mathrm{8}}\left[\left\{\left({tan}^{−\mathrm{1}} \left({tan}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}}\right)\right\}−\left\{\left({tan}^{−\mathrm{1}} \left(\mathrm{0}\right)+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{0}}\right)\right\}\right] \\ $$$${I}=\frac{\pi}{\mathrm{16}}\left[\left({tan}^{−\mathrm{1}} \left(\infty\right)+\frac{\mathrm{1}}{\mathrm{1}+\infty}\right)−\left(\mathrm{1}\right)\right] \\ $$$${I}=\frac{\pi}{\mathrm{16}}\left(\frac{\pi}{\mathrm{2}}+\mathrm{0}−\mathrm{1}\right) \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{32}}−\frac{\pi}{\mathrm{16}} \\ $$$${pls}\:{check}… \\ $$