Question Number 86634 by liki last updated on 29/Mar/20
Answered by MJS last updated on 29/Mar/20
![∫((cos^2 θ)/(tan θ −1))dθ= [t=tan θ → dθ=cos^2 θ dt] =∫(dt/((t−1)(t^2 +1)^2 ))= =−((t−1)/(4(t^2 +1)))+(1/8)ln (((t−1)^2 )/(t^2 +1)) −(1/2)arctan t = =(1/4)cos θ (sin θ +cos θ) +(1/4)ln ∣sin θ −cos θ∣ −(θ/2) +C](https://www.tinkutara.com/question/Q86637.png)
$$\int\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{tan}\:\theta\:−\mathrm{1}}{d}\theta= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\theta\:\rightarrow\:{d}\theta=\mathrm{cos}^{\mathrm{2}} \:\theta\:{dt}\right] \\ $$$$=\int\frac{{dt}}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$=−\frac{{t}−\mathrm{1}}{\mathrm{4}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:{t}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\theta\:\left(\mathrm{sin}\:\theta\:+\mathrm{cos}\:\theta\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{sin}\:\theta\:−\mathrm{cos}\:\theta\mid\:−\frac{\theta}{\mathrm{2}}\:+{C} \\ $$
Commented by liki last updated on 30/Mar/20
$$\boldsymbol{{T}}{hank}\:{you}\:{sir} \\ $$