$\int_0^a \sqrt{1-b^2 x^2} dx$

I'm working on a WKB approximation problem for my quantum mechanics class and came across a weird substitution I would have never thought to make.

The integral I am faced with is:

$$\int_0^a \sqrt{1-b^2 x^2} dx$$

My professor's solution says to make the following substitution.

$$\text{Let} \, \, bx=\sin\theta \Rightarrow dx=\frac{1}{b}\cos\theta d\theta$$

So that $$\int_0^{\arcsin(ba)} \cos^2 (\theta)d\theta.$$

I have never seen a substitution like this, and am worried if I see a similar problem on a test I wouldn't recognize to use it. Are there any other ways to evaluate this integral quickly, or is this the typical way to approach integrals of this form? I'm mainly asking this question to help build my intuition with difficult integrals.

Thank you.

• Yes this is the standard method for such integrals – David Quinn Apr 2 '18 at 16:19
• ... because $\sin^2 + \cos^2 = 1$. (Adding to @DavidQuinn 's comment) – Ethan Bolker Apr 2 '18 at 16:20

4 Answers

If you see $1-y^2$, you should immediately have in mind Pythagorean trigonometric identity. Hence $y=\cos x$ or $y=\sin x$.

Use that $$\cos(2x)=\cos^2(x)-\sin^2(x)=2\cos^2(x)-1$$ For your Control we get $$\frac{a b \sqrt{1-a^2 b^2}+\sin ^{-1}(a b)}{2 b}$$

Hint: Use by integration by parts, use 1 as second function.

By a rescaling of the variable, you can get rid of the parameter $b$. Then the expression $\sqrt{1-x^2}$ reminds the identity $\sqrt{1-\cos^2t}=\pm\sin t$ which allows to remove the square root.

You can also try by parts,

$$I:=\int \sqrt{1-x^2}dx=x\sqrt{1-x^2}+\int\frac{x^2}{\sqrt{1-x^2}}dx.$$

Now you can have the feeling to be in a worse position, but use

$$\int\frac {x^2}{\sqrt{1-x^2}}dx=\int\frac {1-(1-x^2)}{\sqrt{1-x^2}}dx=\int\frac1{\sqrt{1-x^2}}dx-\int\sqrt{1-x^2}dx$$ where the first term is an known derivative ($\arcsin$) and the second is the initial integral.

Here you could also feel to be in a dead-end, but by moving this term to the LHS,

$$2I=x\sqrt{1-x^2}+\arcsin x.$$