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Integral Calculus> Please thiS one fast .......................

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

Mehul Verma , 6 Years ago

Grade 12

1 Answers

Susmita

Last Activity: 6 Years ago

Put

cot^-1(e^x)=z

Or,e^x=cotz

or,e^xdx=-cosec²zdz

Now multiply the numerator and denominator by e^x.

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(e^x)

tanz=1/cotz=1/e^x

$Cos2z=\frac{1-tan^2z}{1+tan^2z}=\frac{e^x-1}{e^x+1}$

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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