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