What is the integral of int(x tan^-1 x)/(1+x^2)^(3/2)∫xtan−1x(1+x2)32 ?
int(x tan^-1 x)/(1+x^2)^(3/2)∫xtan−1x(1+x2)32
1 Answer
Explanation:
I=int(xtan^-1(x))/(1+x^2)^(3/2)color(red)(dx)I=∫xtan−1(x)(1+x2)32dx
Let
Then:
I=int(tanthetatan^-1(tantheta))/(1+tan^2theta)^(3/2)sec^2thetad thetaI=∫tanθtan−1(tanθ)(1+tan2θ)32sec2θdθ
Remember that
I=int(thetatantheta)/sec^3thetasec^2thetad thetaI=∫θtanθsec3θsec2θdθ
I=intthetatantheta 1/secthetad thetaI=∫θtanθ1secθdθ
I=intthetasintheta/costhetacosthetad thetaI=∫θsinθcosθcosθdθ
I=intthetasinthetad thetaI=∫θsinθdθ
This looks primed for integration by parts. Let:
{(u=theta,du=d theta),(dv=sinthetad theta,v=-costheta):}
Then:
I=uv-intvdu
I=-thetacostheta+intcosthetad theta
I=-thetacostheta+sintheta+C
Remember that
Thus
I=-tan^-1(x)/sqrt(x^2+1)+1/sqrt(x^2+1)+C
I=(1-tan^-1(x))/sqrt(x^2+1)+C