Evaluate
∫40lnx√4x−x2dx
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=x−2 this way 4x−x2=4−(x−2)2=4−t2. Substitute,
∫2−2log(t+2)√4−t2 dt
Now let, θ=sin−1t2 so that 2sinθ=t and hence, after substitute,
∫π/2−π/2log[2(1+sinθ)]2cosθ2cosθ dθ=πlog2+∫π/2−π/2log(1+sinθ) dθ
To solve this integral, replace θ by −θ,
I=∫π/2−π/2log(1+sinθ) dθ=∫π/2−π/2log(1−sinθ) dθ
Now,
I+I=∫π/2−π/2log(1−sin2θ) dθ=4∫π/20log(cosθ) dθ
The last integral is a well-known integral that computes to −π2log2.
Your final answer is, πlog2−πlog2.
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