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

Question

I wanted to find $\int\frac{\sqrt{1-t^2}}{1+t^2}dt$, so I substituted $t=\sin\theta$ and got $\int\frac{\cos^2\theta}{1+\sin^2\theta}d\theta$;

but I'm not sure what the best way to proceed from here would be.

Answer 1

$$\int\frac{\cos^2\theta}{1+\sin^2\theta}d\theta$$

multiply num and denum by $\sec^4\theta$

$$\int\frac{\sec^2\theta}{\sec^4\theta+\sec^2\theta\tan^2\theta}d\theta$$

trigonometric identity $\sec^2\theta=1+\tan^2\theta$

$$\int\frac{\sec^2\theta}{2\tan^4\theta+3\tan^2\theta+1}d\theta$$

substitute $u=\tan\theta$

The rest is partial fraction expansion + inverse trigonometric functions

Answer 2

Hint

You arrived to $$I=\int\frac{\cos^2(\theta)}{1+\sin^2(\theta)}d\theta$$ Use the double angle formula to express $\cos^2(\theta)$ and $\sin^2(\theta)$ as function of $\cos (2\theta)$. So, $$I=\int\frac{1+\cos (2\theta)}{3-\cos (2\theta)}d\theta$$ Now, use the tangent half-angle substitution $t=\tan(\theta)$ $$I=\int\frac{2}{2 t^4+3 t^2+1}dt$$ The roots of $2x^2+3x+1=0$ being $-1$ and $-\frac 12$, then continue with partial fraction decomposition $$I=\int\frac{4}{2 t^2+1}dt-\int\frac{2}{t^2+1}dt$$ The second integral does not make any problem and, for the first one, use $t\sqrt 2=y$ to arrive to something similar.