Find the antiderrivative \int 2x ^{\frac{1}{3}} (1-x^{\frac{4}{3}})^2 dx

Question:

Find the antiderrivative {eq}\displaystyle \int 2x ^{\frac{1}{3}} (1-x^{\frac{4}{3}})^2 dx {/eq}

Integration Using Substitution

We use substitution method for sloving the given integral. After substitution we have to integrate the function.

We have used the formulas,

{eq}\displaystyle \frac{dx^n}{dx}=nx^{n-1} {/eq}

{eq}\displaystyle \int x^n dx =\frac{x^{n+1}}{n+1}+c {/eq} (where c is an integrating factor)

Given

{eq}\displaystyle \int 2x ^{\frac{1}{3}} (1-x^{\frac{4}{3}})^2 dx {/eq}

Let us substitute,

{eq}\displaystyle 1-x^{\frac{4}{3}}=t {/eq}

On differentiating,

{eq}\displaystyle -\frac{4}{3}x^{\frac{1}{3}}dx=dt {/eq}

{eq}\displaystyle 2x^{\frac{1}{3}}=\frac{-3}{2}dt {/eq}

The given question changes to,

{eq}\displaystyle \begin{align} \int \frac{-3}{2}t^2dt&=\frac{-3t^3}{2*3}+c\\ &=\frac{-1}{2}( 1-x^{\frac{4}{3}})^3+c \end{align} {/eq}

(where c is an integrating constant)

Therefore

{eq}\displaystyle \int 2x ^{\frac{1}{3}} (1-x^{\frac{4}{3}})^2 dx=\frac{-1}{2}( 1-x^{\frac{4}{3}})^3+c {/eq}