#### Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 59 maths.

Hint: Use indefinite integral formula and the limits to solve this integral

Given: $\int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx$

Solution:

\begin{aligned} &\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x\\ &=\int_{\frac{1}{3}}^{1} \frac{x\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{4}} d x=\int_{\frac{1}{3}}^{1} \frac{\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{3}} d x\ \end{aligned}

Put $\frac{1}{x^2}-1=t\Rightarrow \frac{-2}{x^3}dx=dt\Rightarrow dx=\frac{-x^3}{2}dt$
When $x=\frac{1}{3}$  then t=8 and when x=1 then t=0

\begin{aligned} &\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x=\int_{\frac{1}{3}}^{1} \frac{\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{3}} d x \\ &=-\int_{8}^{0} \frac{t^{\frac{1}{3}}}{2} d t \\ &=-\frac{1}{2} \int_{8}^{0} t^{\frac{1}{3}} d t=-\frac{1}{2}\left[\frac{t^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{8}^{0}=-\frac{1}{2}\left[\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]_{8}^{0} \\ &=-\frac{1}{2} \times \frac{3}{4}\left[t^{\frac{4}{3}}\right]_{8}^{0}=\frac{-3}{8}\left[0^{\frac{4}{3}}-8^{\frac{4}{3}}\right] \\ &=\frac{-3}{8}\left[0-2^{4}\right]=\frac{-3}{8}(-16)=2 \times 3=6 \end{aligned}