# Find $\int \dfrac{dt}{t-\sqrt{1-t^2}}$

Find $$\int \dfrac{dt}{t-\sqrt{1-t^2}}$$

MY APPROACH :

1. Substitute $t = \sin x$

2. Multiply numerator and denominator by $\cos x+\sin x$

then rewrite everything in terms in $\sin2x$ and $\cos2x$, we get Integrable functions.

But may be there is a better way. Can anyone think of anything smarter, thanks :)

\begin{align}\int\dfrac{dt}{t-\sqrt{1-t^2}} &\stackrel{\small t=\sin x}\equiv\int\frac{\cos xdx}{\sin x-\cos x}\\ &=\frac12\left(\int\frac{\cos x+\sin x}{\sin x-\cos x}dx-\int\frac{\sin x-\cos x}{\sin x-\cos x}dx\right)\\ &=\frac12(\ln(t-\sqrt{1-t^2})-\arcsin t)+C\end{align}

• How do you get to the third equality? – UserX Sep 15 '14 at 16:59
• @UserX $d(\sin x-\cos x)=(\cos x+\sin x)dx$ – RE60K Sep 15 '14 at 17:00

Faster approach;

$$u=\sin t \cdots$$ $$\int \frac{\cos u }{\sin u- \cos u} \mathrm{d}u$$

Multiply the integrand by $$\frac{\sec^3(u)}{\sec^3(u)}$$

Then you got $$\int \frac{\sec^2(u)}{\sec^2(u) \tan(u)-\sec^2(u)} \mathrm{d}x \stackrel{\sec^2(u)=\tan^2(u)+1}{=}\int \frac{\sec^2(u)}{(\tan(u)-1) (1+\tan^2(u))} \mathrm{d}x$$

Substitute $s=\tan(u)$

Then use partial fraction decomposition.

Enjoy.

• Faster wrt?${}{}$ – RE60K Sep 15 '14 at 16:34
• I was expecting this "faster wrt OP's approach" ? – RE60K Sep 15 '14 at 16:49
