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Question Number 183123 by SANOGO last updated on 21/Dec/22

$$\int ln\left(tanx\right)dx$$

Answered by TheSupreme last updated on 21/Dec/22

$$\int ln\left(\frac{sin\left(x\right)}{cos\left(x\right)}\right)dx$$$$t=tan\left(x\right)$$$$\frac{dt}{1+t^2}=dx$$$$\int \frac{ln\left(t\right)}{1+t^2}dt=ln\left(t\right)arctan\left(t\right)-\int \frac{1}{t(1+t^2)}dt=$$$$ln\left(t\right)arctan\left(t\right)-\int \frac{A}{t}+\frac{Bt+C}{1+t^2}dt$$$$A+{At}^2+{Bt}^2+Ct=1$$$$A=1$$$$C=0$$$$B=-1$$$$I=ln\left(t\right)arctan\left(t\right)-\int \frac{1}{t}dt+\int \frac{t}{1+t^2}dt$$$$I=ln\left(t\right)arctan\left(t\right)-ln\left(t\right)+\frac{1}{2}ln(1+t^2)$$$$I=ln\left(tan\left(x\right)\right)(arctan\left(tan\left(x\right)\right)-1)-ln\left(cos\left(x\right)\right)$$

Commented by MJS_new last updated on 21/Dec/22

$$\mathrm{wrong}.\mathrm{why}\mathrm{does}\mathrm{noone}\mathrm{test}\mathrm{their}\mathrm{results}\mathrm{anymore}?$$$$\int \frac{\ln t}{t^2+1}dt=$$$$u^{\prime }=\frac{1}{t^2+1}\to u=\arctan t$$$$v=\ln t\to v^{\prime }=\frac{1}{t}$$$$\int u^{\prime }v=uv-\int {uv}^{\prime }$$$$=\ln t\arctan t-\int \frac{\arctan t}{t}dt$$

Commented by SANOGO last updated on 21/Dec/22

$$mercibien$$

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