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Question Number 183123 by SANOGO last updated on 21/Dec/22
∫ln(tanx)dx
Answered by TheSupreme last updated on 21/Dec/22
∫ln(sin(x)cos(x))dxt=tan(x)dt1+t2=dx∫ln(t)1+t2dt=ln(t)arctan(t)−∫1t(1+t2)dt=ln(t)arctan(t)−∫At+Bt+C1+t2dtA+At2+Bt2+Ct=1A=1C=0B=−1I=ln(t)arctan(t)−∫1tdt+∫t1+t2dtI=ln(t)arctan(t)−ln(t)+12ln(1+t2)I=ln(tan(x))(arctan(tan(x))−1)−ln(cos(x))
Commented by MJS_new last updated on 21/Dec/22
wrong.whydoesnoonetesttheirresultsanymore?∫lntt2+1dt=u′=1t2+1→u=arctantv=lnt→v′=1t∫u′v=uv−∫uv′=lntarctant−∫arctanttdt
Commented by SANOGO last updated on 21/Dec/22
mercibien
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