Question Number 220676 by fantastic last updated on 17/May/25
$$\int\:\frac{{xdx}}{\left(\mathrm{1}−{cosx}\right)^{\mathrm{2}} } \\ $$
Answered by Frix last updated on 17/May/25
![I=∫(x/((1−cos x)^2 ))dx =^([by parts]) =−(((2−cos x)(1+cos x)^2 x)/(3sin^3 x))+∫(((2−cos x)(1+cos x)^2 )/(3sin^3 x))dx ∫(((2−cos x)(1+cos x)^2 )/(3sin^3 x))dx =^(t=cos x) =(1/3)∫((t−2)/((t−1)^2 ))dt=((ln ∣t−1∣)/3)+(1/(3(t−1)))= =((ln ∣1−cos x∣)/3)−(1/(3(1−cos x))) ⇒ I=−(((2−cos x)(1+cos x)x+sin x)/(3(1−cos x)sin x))+((ln ∣1−cos x∣)/3)+C](https://www.tinkutara.com/question/Q220685.png)
$${I}=\int\frac{{x}}{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{dx}\:\overset{\left[\mathrm{by}\:\mathrm{parts}\right]} {=} \\ $$$$=−\frac{\left(\mathrm{2}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} {x}}{\mathrm{3sin}^{\mathrm{3}} \:{x}}+\int\frac{\left(\mathrm{2}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{\mathrm{3sin}^{\mathrm{3}} \:{x}}{dx} \\ $$$$ \\ $$$$\int\frac{\left(\mathrm{2}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{\mathrm{3sin}^{\mathrm{3}} \:{x}}{dx}\:\overset{{t}=\mathrm{cos}\:{x}} {=} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{t}−\mathrm{2}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt}=\frac{\mathrm{ln}\:\mid{t}−\mathrm{1}\mid}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}\left({t}−\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{ln}\:\mid\mathrm{1}−\mathrm{cos}\:{x}\mid}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)} \\ $$$$\Rightarrow \\ $$$${I}=−\frac{\left(\mathrm{2}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}\right){x}+\mathrm{sin}\:{x}}{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)\mathrm{sin}\:{x}}+\frac{\mathrm{ln}\:\mid\mathrm{1}−\mathrm{cos}\:{x}\mid}{\mathrm{3}}+{C} \\ $$