Using a substitution of a square of a trigonometric function.

Question

Are there any cases where we could use a substitution, for example, $x=\tan^2(\theta)$? If so, how would one go about it? I tried to solve a simple integral $\int \sqrt{x-1}dx $ with the substitution $x=\sec^2{\theta}$. However, i ended up with $2/3\tan^3{\theta}$ which i have no idea how to transform back to $x$ except getting $2/3\tan{\theta}(x-1)$.

Thanks!

Answer 1

If you have $x=\sec^2(\theta)$ then $\tan(\theta)=\sqrt{x-1}$ or its negative and thus on the positive side $$\tfrac23 \tan^3(\theta)=\tfrac23 (x-1)^{3/2}$$ Indeed $$\int \sqrt{x-1}\,dx = \tfrac23 (x-1)^{3/2} +C$$ so your method works, though you do not really need substitution for this

Answer 2

Since

$$x=\sec^2{\theta} \implies dx =2 \tan\theta \sec^2 \theta \,d\theta$$

$$\sqrt{x-1}=\sqrt{\sec^2{\theta}-1}=\tan \theta$$

$$\int \sqrt{x-1}dx=2\int \tan^2\theta\sec^2 \theta \,d\theta=\frac23\tan^3\theta$$

and

$$\tan(\arccos x) =\frac{\sqrt{1 - x^2}}{x}$$

$$\theta=\arccos \frac{1}{\sqrt x}$$

we have that

$$\int \sqrt{x-1}dx=\frac23\tan^3\theta=\frac23 \left(\frac{\sqrt{1 - \frac1x}}{\frac{1}{\sqrt x}}\right)^3=\frac23(x-1)^\frac32$$