Thursday, January 19, 2017

calculus - Evaluate int40fraclnxsqrt4xx2,mathrmdx




Evaluate
40lnx4xx2dx




How do I evaluate this integral? I know that the result is 0, but I don't know how to obtain this. Wolfram|Alpha yields a non-elementary antiderivative for the indefinite integral, so I don't think I can directly integrate and then plug in the upper/lower limits.


Answer




First let t=x2 this way 4xx2=4(x2)2=4t2. Substitute,
22log(t+2)4t2 dt
Now let, θ=sin1t2 so that 2sinθ=t and hence, after substitute,
π/2π/2log[2(1+sinθ)]2cosθ2cosθ dθ=πlog2+π/2π/2log(1+sinθ) dθ
To solve this integral, replace θ by θ,
I=π/2π/2log(1+sinθ) dθ=π/2π/2log(1sinθ) dθ
Now,
I+I=π/2π/2log(1sin2θ) dθ=4π/20log(cosθ) dθ
The last integral is a well-known integral that computes to π2log2.



Your final answer is, πlog2πlog2.



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