Integrate $\frac{1}{x\sqrt{a - bx^2}}$

Question

I've been toiling for hours trying to answer this one:

Find $\int \frac{1}{x\sqrt{a - bx^2}}\,dx$

It's from Silvanus Thompson's 'Calculus Made Easy' (apparently not easy enough for me!).

The answer given is $\frac {1}{\sqrt{a}}\log\left(\frac{\sqrt{a}-\sqrt{a-bx^2}}{x\sqrt{a}}\right)$

I've attempted the question by letting $x=\sqrt{\frac ab}\sin(u)$, which I think is on the right track as I end up with the $\frac1{\sqrt a}$ term in my answer, but nothing else falls into place in terms of the given answer.

Thank you in advance for any help!

Andrew

Answer 1

Hint Your method seems to work, provided anyway that $a, b > 0$: The substitution $x = \sqrt{\frac{a}{b}} \sin \theta$, $dx = \sqrt{\frac{a}{b}} \cos \theta \,d\theta$ transforms the integral to $$\frac{1}{\sqrt{a}} \int \csc \theta \,d\theta .$$

Another option avoids trigonometric functions altogether: Rewrite the integral as $$\int \frac{x \,dx}{x^2 \sqrt{a - b x^2}} .$$

What substitution does this form of the integrand suggest?

Additional hint Consider the substitution $u^2 = a - b x^2$, $du = -2 b x \,dx$.

Answer 2

$$I=\int \frac{dx}{x\sqrt{a-bx^2}}$$ Let $x=\sqrt\frac{a}{b}\sin t$, then $$I=\int \frac{\sqrt{a/b} \cos t!dt}{\sqrt{a/b} \sin t \sqrt{a} \cos t}.$$ $$\implies I=\frac{1}{\sqrt{a}} \int \csc t dt=\frac{1}{\sqrt{a}}\int \csc t \frac{\csc t+\cot t}{\csc t+ \cot t}dt.$$ Let $\csc t+ \cot t=u$ $$I=\frac{1}{\sqrt{a}} \int \frac{-du}{u}=\frac{-1}{\sqrt{a}}\ln(\csc t+\cot t)=-\frac{1}{\sqrt{a}}ln\frac{1-\sqrt{1-c^2x^2}}{cx}+C,c=\sqrt{b/a}$$

Answer 3

Hint:

$$\int\dfrac{dx}{x\sqrt{a-bx^2}}=\int\dfrac{bx\ dx}{bx^2\sqrt{a-bx^2}}$$

Let $\sqrt{a-bx^2}=u\implies du=\dfrac{{-bx}}{\sqrt{a-bx^2}}$ and $bx^2=a-u^2$

More generally for $$\int\dfrac{dx}{x\sqrt{a-bx^n}}=\int\dfrac{bx^{n-1}\ dx}{bx^n\sqrt{a-bx^n}}$$

set $\sqrt{a-bx^n}=y$

Answer 4

Substitute $x=\sqrt{\frac ab }\frac1t$ to get

$$\int \frac{dx}{x\sqrt{a - bx^2}}=-\frac1{\sqrt a}\int \frac{dt}{\sqrt{t^2-1}}= - \frac1{\sqrt a}\cosh^{-1}t=- \frac1{\sqrt a}\ln\left(t+\sqrt{t^2-1}\right) $$

which is equal to $\frac {1}{\sqrt{a}}\ln\left(\frac{\sqrt{a}-\sqrt{a-bx^2}}{x\sqrt{a}}\right)$.