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

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.

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

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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.

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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


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


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

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.

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