Proof that: ∫0π2ln⁡(1+αsin2x)dx=πln⁡1+1+α2

Question

Proof that:  ∫0π2ln⁡(1+αsin2x)dx=πln⁡1+1+α2

Carl Swisher · Accepted Answer

Let I(α)=∫0π2ln⁡(1+αsin2x)dxThen differentiating under the integral sign,I′(a)=∫0π2sin2x1+asin2xdx=∫0π21a+csc2xdxNow let u=cot⁡xThenI′(a)=∫0∞1a+1+u211+u2du=1a∫0∞(11+u2−11+a+u2)du=1a(π2−11+a∫0∞11+u21+adu)=1a(π2−11+a∫0∞11+v2dv)=π2a(1−11+a)Then integrating back,I(α)=π2∫1a(1−11+a)da=π2∫1u2−1(1−1u)2udu=π∫11+udu=πln⁡(1+1+a+CAnd since I(0)=0, C=−πln⁡2Therefore,I(a)=πln⁡(1+1+a2)

Wendy Boykin · Accepted Answer

This is quite similar to Random Variable&#039;s solution, just the starting integral is different to make the calculations a bit simpler.ConsiderI(b)=∫0π2ln⁡(b2+sin2x)dx⇒I′(b)=∫0π22bb2+sin2xdx=2b∫0π2dxb2+cos2xFactor out cos2x from the denominator and rewrite sec2x=1+tan2x to obtain:I′(b)=2b∫0π2sec2xdxb2+1+b2tan2xdxUse the substitution tan⁡x=t and evaluating the resulting integral is easy soI′(b)=π1+b2⇒I(b)=πln⁡(b+1+b2)+CFor b=0, I(0)=−πln⁡2, hence C=−πln⁡2⇒∫0π2ln⁡(b2+sin2x)dx=πln⁡(b+1+b22)Replace b with 1α and you get:∫0π2ln⁡(1+αsin2x)dx−π2ln⁡α=πln⁡(1+1+α2α)⇒∫0π2ln⁡(1+αsin2x)dx=πln⁡(1+1+α2)

star233 · Accepted Answer

∫0π/2ln⁡(1+αsin2⁡(x))dx=πln⁡(1+1+α2)&amp;nbsp;ddx∫0π/2ln⁡(1+αsin2⁡(x))dx=ddα∫0π/2ln⁡(1+αcos2⁡(x))dx&amp;nbsp;=∫0π/2cos2⁡(x)1+αcos2⁡(x)dx&amp;nbsp;=1α∫0π/2[1+αcos2⁡(x)]−11+αcos2⁡(x)dx&amp;nbsp;=π2α−∫0π/2dx1+αcos2⁡(x)&amp;nbsp;=π2α−1α∫0π/2sec2⁡(x)dxtan2⁡(x)+1+α&amp;nbsp;=π2(1α−1α1+α)&amp;nbsp;∫0π/2ln⁡(1+αsin2⁡(x))dx&amp;nbsp;Set&amp;nbsp;x≡1+1+t⇒t≡x2−2x&amp;nbsp;=π2∫21+1+a2dxx&amp;nbsp;∫0π/2ln⁡(1+αsin2⁡(x))dx=πln⁡(1+1+α2)

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

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New Questions in Calculus 1