Question

Evaluate: $\int \frac{t^{4}dt}{\sqrt{1-t^2}}$

Answer 1

$\int \frac{t^{4}dt}{\sqrt{1-t^2}}$

Say $t=\sin \theta \Rightarrow dt=\cos \theta d\theta$

$=\int \frac{{\sin }^4\theta }{\sqrt{1-{\sin }^2\theta }}\cos \theta d\theta$

$=\int \frac{{\sin }^4\theta }{{\sqrt{\cos }}^2\theta }\cdot \cos \theta d\theta =\int \frac{{\sin }^4\theta }{\cos \theta }\cos \theta d\theta$

$=\int {\sin }^4\theta d\theta$

$=\frac{1}{4}\int (2{\sin }^2\theta {)}^{2}d\theta =\frac{1}{4}\int (1-\cos 2\theta {)}^{2}d\theta$

$=\frac{1}{4}[\int (1+{\cos }^{2}2\theta -2\cos 2\theta \left)d\theta \right]$

$=\frac{1}{4}[\int d\theta +\int {\cos }^{2}2\theta d\theta -2\int \cos 2\theta d\theta ]$

$=\frac{1}{4}[\int d\theta +\frac{1}{2}\int 2{\cos }^{2}2\theta d\theta -2\int \cos 2\theta d\theta ]$

$=\frac{1}{4}[\int d\theta +\frac{1}{2}(1+\cos 4\theta )d\theta -2\int \cos 2\theta d\theta ]$

$=\frac{!}{4}[\int d\theta +\frac{1}{2}\int d\theta +\frac{1}{2}\cos 4\theta d\theta -2\int \cos 2\theta d\theta ]$

$=\frac{1}{4}[\theta +\frac{\theta }{2}+\frac{\sin 4\theta }{8}-\sin 2\theta ]$

$=\frac{3\theta }{8}+\frac{\sin 4\theta }{32}-\frac{\sin 2\theta }{4}$

$=\frac{3{\sin }^{-1}t}{8}+\frac{4\sin \theta \cdot \cos \theta \cdot (1-2{\sin }^2\theta )}{32}-\frac{2\sin \theta \cdot \cos \theta }{4}$

$=\frac{3}{8}{\sin }^{-1}t+\frac{1}{8}t\sqrt{1-t^2}(1-2+2t^2)-\frac{1}{2}t\sqrt{1-t^2}$

$=\frac{3}{8}{\sin }^{-1}t+\frac{1}{8}t\sqrt{1-t^2}(2t^2-1)-\frac{1}{2}t\sqrt{1-t^2}$