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Please thiS one fast .....................................

Mehul Verma , 6 Years ago
Grade 12
anser 1 Answers
Susmita

Last Activity: 6 Years ago

Put
cot-1(ex)=z
Or,ex=cotz
or,exdx=-cosec2zdz
Now multiply the numerator and denominator by ex.
It will become
\frac{e^x cot^-1(e^x)}{(e^x)^2}
Now replace all x,dx by the values of z,dz.It will become
\frac{-zcosec^2zdz}{cot^2z}
Now we shall carry on integration by parts taking z as 1st function.
-\int zsec^2zdz
=-zsec^2zdz
=-ztanz+logcosz
Now we have to replace z,tanz,cosz in terms of x.
z=cot-1(ex)
tanz=1/cotz=1/ex
Cos2z=\frac{1-tan^2z}{1+tan^2z}=\frac{e^x-1}{e^x+1}
Also we know that
1+cos2z=2cos2z
or,cos2z=(cos2z+1)/2
So, cosz=\sqrt{\frac{e^x}{e^x+1}}
So, logcosz=log\sqrt{\frac{e^x}{e^x+1}}=\frac{1}{2} [loge^x-log(e^x +1)] =\frac{1}{2}[x-log(e^x+1)]Put these all together and the answer will become
\frac{-cot^-1 (e^x)}{x}+\frac{1}{2}x -\frac{1}{2} log (e^x+1)

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