Question

Evaluate the integral: $\int_{0}^{4} \sin^{-1} \frac{2x}{1+x^2}\,dx$

Answer 1

Let $I$ =$\int_{0}^{4} \sin^{-1} \frac{2x}{1+x^2}\,dx$

Also ,let x=$\tan \theta$ $⇒$ dx=$\sec^2 \theta d\theta$

When x=0, $\theta$ =0 and when x=1, $\theta$ =$\frac {π}{4}$

$I$=$\int^{\frac{\pi}{4}}_0 \sin^{-1}$ \frac{(2tanθ}{1+tan^2θ)}$$\sec^2\theta d\theta

=$\int^{\frac{\pi}{4}}_0 \sin^{-1}(\sin2\theta)\sec^2 \theta d \theta$

=$\int^{\frac{\pi}{4}}_0 2\theta.\sec^2 \theta d\theta$

=$2\int^{\frac{\pi}{4}}_0 \theta.\sec^2 \theta d\theta$

Taking θ as first function and sec2 $\theta$ as second function and integrating by parts, we obtain

I=2\bigg[\theta\int\sec^2 \theta d\theta-\int\bigg\\{\bigg(\frac{d}{dx}\theta\bigg)\int\sec^2\theta d\theta\bigg\\}d \theta\bigg]^{\frac{\pi}{4}}_0

=$2\bigg[\theta \tan \theta-\int\tan \theta d\theta\bigg]^{\frac{\pi}{4}}_0$

=$2[\theta \tan \theta+\log|\cos \theta|]^{\frac{\pi}{4}}_0$

=$2\bigg[\frac{\pi}{4}\tan\frac{\pi}{4}+\log|\cos \frac{\pi}{4}|-\log|\cos 0|\bigg]$

=$2\bigg[\frac{\pi}{4}+\log\bigg(\frac{1}{\sqrt2}\bigg)-\log1\bigg]$

=$2\bigg[\frac{\pi}{4}-\frac{1}{2}\log2\bigg]$

=$\frac {π}{2}$-log 2