# Answer to Question #20341 in Calculus for sakura

Question #20341

can you please help me this integral of [arccot sqrt(z)]/sqrt(z)?

Expert's answer

From the definition, integral of arccotangent isz*arccot(z)+0.5ln(1+z^2)

To evaluate this integral we should change variables:t=sqrt(z)

So, dt=1/2sqrt(z)dz

Arccot(sqrt(z)/sqrt(z)dz=2arccot(t)dt

Integrating this we get:2*t*arccot(t)+ln(1+t^2)=2*sqrt(z)*arccot(sqrt(z))+ln(1+z)

To evaluate this integral we should change variables:t=sqrt(z)

So, dt=1/2sqrt(z)dz

Arccot(sqrt(z)/sqrt(z)dz=2arccot(t)dt

Integrating this we get:2*t*arccot(t)+ln(1+t^2)=2*sqrt(z)*arccot(sqrt(z))+ln(1+z)

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