# Calculate $\int (6x^2-2)^{\frac{3}{2}} \mathrm{d}x$

$$\int (6x^2-2)^{\frac{3}{2}} \mathrm{d}x$$

Tried converting to trigonometric functions using substitution $$6x^2-2 = t^2$$ and then $$t^2 = 2\tan \theta$$, but I get an equation in $$\sec \theta$$ with higher powers like $$\int \sec^5 \theta$$ etc. How do I solve these or the original problem, any hints would be helpful. Thanks.

• $\int \sec^5 t \mathrm dt$ could be calculated via establishing recurrence relations. Note that $\mathrm d(\tan t) = \sec^2 t.$ – xbh Oct 11 '19 at 3:17
• can you elaborate on recurrence relations? – Rick Oct 11 '19 at 3:28

The standard reduction formula for $$\int \sec^n \theta \,d\theta , \qquad n \geq 3,$$ is $$\boxed{\int \sec^n \theta \,d\theta = \frac{1}{n-1} \sec^{n-2} \theta \tan \theta + \frac{n-2}{n-1} \int \sec^{n - 2} \theta \,d\theta} .$$ (See this question for more information about this formula, and see either of its answers for a derivation.)

Substituting $$n = 5$$ then gives $$\int \sec^5 \theta \,d\theta = \frac{1}{4}\sec^3 \theta \tan \theta + \frac{3}{4} \int \sec^3 \theta \,d\theta .$$ Applying the same rule to $$\int \sec^3 \theta \,d\theta$$ gives an expression for $$\int \sec^5 \theta \,d\theta$$ in terms of $$\int \sec \theta \,d\theta$$ and closed-form expressions, and that latter integral can be evaluated in many ways.

Rewrite the integral as,

$$I=\int (6x^2-2)^{\frac{3}{2}} dx =\sqrt{\frac 83} \int [(\sqrt 3 x)^2-1]^{3/2}d(\sqrt 3 x)$$

and let $$\sqrt 3 x=\cosh t$$,

$$I=\sqrt{\frac 83} \int \sinh^4 t dt$$

Expend the integrand with the identities $$2\sinh^2 t = \cosh 2t -1$$ and $$2\cosh^2 2t = \cosh 4t +1$$,

$$\sinh^4 t = \frac 38 - \frac 12 \cosh 2t +\frac 18 \cosh 4t$$

Then, integrate,

$$I=\sqrt{\frac 23}\left(\frac 34 t -\frac12 \sinh 2t+ \frac{1}{16}\sinh 4t\right)+C$$

Note that the substitution with $$\cosh t$$ results in economical integration afterwards.

First of all take $$2^{\frac{3}{2}}$$ out of the expression

Then use $$x=\frac{\sec(u)}{\sqrt{3}}$$

In the end you'll get

$$\frac{2^{\frac{3}{2}}}{\sqrt{3}}\int \sec(u) {\tan (u)}^4 \ du$$

If you solve this and substitute back you'll get

$$\dfrac{\sqrt{3}\ln\left(\left|\sqrt{3x^2-1}+\sqrt{3}x\right|\right)+x\sqrt{3x^2-1}\left(6x^2-5\right)}{2^\frac{3}{2}}$$

• and how does one solve that? any hints? Like that is what I got and how do I proceed from there? – Rick Oct 11 '19 at 5:08
• You could use integral calculator with steps online to see all the steps. – Harshit Gupta Oct 11 '19 at 5:09
• I think,here,showing all the steps is too much for a question – Harshit Gupta Oct 11 '19 at 5:09
• Nevermind I got the relevant hint. Thanks anyway. – Rick Oct 11 '19 at 5:16

Let

$$I=\int (6x^2-2)^{\frac{3}{2}} \mathrm{d}x$$

where we perform the substitution $$x=\dfrac{\sec u}{\sqrt{3}}$$. Then, $$dx=\dfrac{\tan u \sec u}{\sqrt{3}}\,du$$ and

$$(6x^2-2)^{3/2} =\left(2\sec^2{u}-2\right)^{3/2}=2\sqrt{2}\tan^3{u}$$

therefore as $$u=\sec^{-1}\left(\sqrt{3x}\right)$$ we have that

$$I=\frac{2\sqrt{2}}{\sqrt{3}}\int\tan^4 u \sec u\,du$$

which we integrate by parts with $$\tilde{u}=\tan^3 u$$ and $$dv=\sec{u}\tan{u}$$ to form

$$I=\frac{2\sqrt{2}}{\sqrt{3}}\left(\sec u\tan^3 u-3I-3\int\tan^2u\sec u \,du\right)$$

suppose now that

$$J=\int\tan^2u\sec u \,du$$

integrating by parts again with $$\tilde{u}=\tan u$$ and $$dv=\sec{u}\tan{u}$$ forms

$$J=\sec{u}\tan{u}-J-\ln|\sec u + \tan u|$$

which can be rewritten as

$$J=\frac 12\Big(\sec{u}\tan{u}-\ln|\sec u + \tan u| \Big)+C$$

so that

$$I=\frac{1}{\sqrt{6}}\left(\sec u\tan^3 u-\frac 32\Big(\sec{u}\tan{u}-\ln|\sec u + \tan u| \Big)\right)+C$$

where we now perform the original substitution $$u=\sec^{-1}\left(\sqrt{3x}\right)$$ to obtain a closed form expression.