# Integral $\int\frac{\sqrt{4x^2-1}}{x^3}dx$ using trig identity substitution!

$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$

So, make the substitution

$$x = \sqrt{a \sec \theta}$$, which simplifies to $$a \tan \theta$$.

$$2x = \sqrt{1} \sec \theta$$,

$$d\theta = \dfrac{\sqrt{1}\sec\theta\tan\theta}{2}$$

$$\int \dfrac{\sqrt{1}\tan\theta}{(\sqrt{1}\sec\theta)^3} d\theta$$

Am I making the correct substitutions here? Substituting $$d\theta$$ a quantity of $$(\sqrt{1}\sec\theta)$$ will cancel from the denominator. Somewhere along the line I need to use the identity $$\sin(2\theta)=2\sin(\theta)\cos(\theta).$$

• Check again, the result is $4\int\sin^2\theta\ d\theta$. – Nosrati Oct 3 '18 at 2:54
• Care to explain how $x = \sqrt{a}\sec\theta$ gets me to that point? – DJ2 Oct 3 '18 at 3:24
• It is $x=\sqrt{a}\sec\theta$. – Nosrati Oct 3 '18 at 3:26
• Ah, yes @Nosrati – DJ2 Oct 3 '18 at 3:28

With the substitution $$x=\frac {\sec \theta }{2}$$ you get $$dx = \frac {\sec \theta \tan \theta }{2} d\theta$$ and the integral changes to $$\int \frac {4\tan^2 \theta \sec \theta }{ \sec ^3 \theta } d\theta =4 \int \frac {\tan^2 \theta }{ \sec ^2 \theta } d\theta =4 \int \sin ^2 \theta d\theta$$

Now you can use the double angle equality which you mentioned.

• Should be: $dx = \frac{1}{2} \sec \theta \tan \theta d \theta$. – JavaMan Oct 3 '18 at 3:19
• Can you explain what to do after substituting the double angle identity? – DJ2 Oct 3 '18 at 3:34
• $\sin ^2 \theta =(1-\cos( 2\theta ))/2$ and integrate the result. – Mohammad Riazi-Kermani Oct 3 '18 at 10:39

Take $$2x = \cos(\theta)$$, then you will get

$$\int\frac{\sqrt{\cos^2(\theta)-1}}{\cos^3(\theta)} d\theta= \int i\frac{\sin(\theta)}{\cos^3(\theta)} d\theta$$

then take another substitution. $$t=\cos(\theta)$$ to finally get : $$=\frac{i}{8x^2} + c$$

• $\cos^2-1\neq\sin^2$. – Nosrati Oct 3 '18 at 3:07
• @Nosrati. Yes you are right. Typo on my side. Thanks for notifying. Edit made – Kashan Oct 3 '18 at 3:18
• Presumably this is question is for a calculus course, so we are (probably) tacitly assuming all quantities should be real. – JavaMan Oct 3 '18 at 3:20
• @JavaMan Correct. – DJ2 Oct 3 '18 at 3:21

Hint:

If trigonometric substitution is not mandatory, for (where $$a$$ is an arbitrary constant )

$$\dfrac{\sqrt{4x^2+a}}{x^{2n-1}}=4^{n-1}\dfrac{\sqrt{4x^2+a}}{(4x^2)^n}\cdot4x$$

set $$\sqrt{4x^2+a}=y,4x^2+a=y^2,y\ dy=4x\ dx$$