Proof that: \int_0^{\pi/2}\ln(1+\alpha\sin^2 x)dx=\pi\ln\frac{1+\sqrt{1+\alpha}}{2}

Tara Alvarado

Tara Alvarado

Answered question

2022-01-03

Proof that:
0π2ln(1+αsin2x)dx=πln1+1+α2

Answer & Explanation

Carl Swisher

Carl Swisher

Beginner2022-01-04Added 28 answers

Let I(α)=0π2ln(1+αsin2x)dx
Then differentiating under the integral sign,
I(a)=0π2sin2x1+asin2xdx=0π21a+csc2xdx
Now let u=cotx
Then
I(a)=01a+1+u211+u2du
=1a0(11+u211+a+u2)du
=1a(π211+a011+u21+adu)
=1a(π211+a011+v2dv)
=π2a(111+a)
Then integrating back,
I(α)=π21a(111+a)da
=π21u21(11u)2udu
=π11+udu
=πln(1+1+a+C
And since I(0)=0, C=πln2
Therefore,
I(a)=πln(1+1+a2)
Wendy Boykin

Wendy Boykin

Beginner2022-01-05Added 35 answers

This is quite similar to Random Variable's solution, just the starting integral is different to make the calculations a bit simpler.
Consider
I(b)=0π2ln(b2+sin2x)dx
I(b)=0π22bb2+sin2xdx=2b0π2dxb2+cos2x
Factor out cos2x from the denominator and rewrite sec2x=1+tan2x to obtain:
I(b)=2b0π2sec2xdxb2+1+b2tan2xdx
Use the substitution tanx=t and evaluating the resulting integral is easy so
I(b)=π1+b2I(b)=πln(b+1+b2)+C
For b=0, I(0)=πln2, hence C=πln2
0π2ln(b2+sin2x)dx=πln(b+1+b22)
Replace b with 1α and you get:
0π2ln(1+αsin2x)dxπ2lnα=πln(1+1+α2α)
0π2ln(1+αsin2x)dx=πln(1+1+α2)
star233

star233

Skilled2022-01-11Added 403 answers

0π/2ln(1+αsin2(x))dx=πln(1+1+α2) ddx0π/2ln(1+αsin2(x))dx=ddα0π/2ln(1+αcos2(x))dx =0π/2cos2(x)1+αcos2(x)dx =1α0π/2[1+αcos2(x)]11+αcos2(x)dx =π2α0π/2dx1+αcos2(x) =π2α1α0π/2sec2(x)dxtan2(x)+1+α =π2(1α1α1+α) 0π/2ln(1+αsin2(x))dx Set x1+1+ttx22x =π221+1+a2dxx 0π/2ln(1+αsin2(x))dx=πln(1+1+α2)

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