Evaluate \int \frac{3x^2}{\sqrt{1-x^6}} \mathrm{d}x .

Question:

Evaluate

{eq}\int \frac{3x^2}{\sqrt{1-x^6}} \mathrm{d}x {/eq}.

Integration:

Integration is an important part of the calculus. Integration is used to calculate area, volume, etc.

Here, we have to evaluate the integral of a given function.

There are many ways to evaluate the integral like substitution method, integration by parts, integration by partial fraction, etc. But given problem is in the form of {eq}\int f(g(x))g'(x)dx {/eq}, so here we use the substitution method to evaluate the integral.

Given

{eq}\int \dfrac{3x^2}{\sqrt{1-x^6}} \mathrm{d}x {/eq}.

We have to solve the integral.

Here, we use the substitution {eq}x^3=t {/eq} to solve the integral.

Differentiate both sides of the substitution with respect to {eq}x {/eq}.

{eq}\begin{align} x^3 &=t\\ \dfrac{\mathrm{d} }{\mathrm{d} x}(x^3) &=\dfrac{\mathrm{d} t}{\mathrm{d} x}\\ 3x^2 &=\dfrac{\mathrm{d} t}{\mathrm{d} x}\\\ (3x^2)dx &=dt \end{align} {/eq}

Applying the substitution to the integral, we have:

{eq}\begin{align} \int \dfrac{3x^2}{\sqrt{1-x^6}} \mathrm{d}x &=\int \dfrac{dt}{\sqrt{1-t^2}}\\ &=\arcsin(t)+C\ & \left [ \because \int \dfrac{1}{\sqrt{a^2-x^2}}=\arcsin\left ( \dfrac{x}{a} \right ) \right ] \end{align} {/eq}

Reversing the substitution, we have:

{eq}\color{blue}{\int \dfrac{3x^2}{\sqrt{1-x^6}} =\arcsin(x^3)+C} {/eq}