Question

The value of the integral ${\displaystyle {\int }_{\frac{1}{3}}^1\frac{{(x-x^3)}^{\frac{1}{3}}}{x^4} dx}$ is

विकल्प

• 6

• 0

• 3

• 4

Answer 1

6

Explanation:

Let I = ${\displaystyle {\int }_{\frac{1}{3}}^1 \frac{{(x-x^3)}^{\frac{1}{3}}}{x^4}}$

Put ${\displaystyle x=\sin \theta ,dx=\cos \theta d\theta }$

When ${\displaystyle x=\frac{1}{3},\theta ={\sin }^{-1} \frac{1}{3}}$, When ${\displaystyle x=1,\theta =\frac{\pi }{2}}$

∴ I = ${\displaystyle {\int }_{{\sin }^{-1} \frac{1}{3}}^{\frac{\pi }{2}} \frac{{(\sin \theta -{\sin }^3\theta )}^{\frac{1}{3}}}{{\sin }^4\theta }\cos \theta d\theta }$

= ${\displaystyle {\int }_{{\sin }^{-1} \frac{1}{3}}^{\frac{\pi }{2}}\frac{{[\sin \frac{1}{3} \theta (1-{\sin }^2\theta )]}^{\frac{1}{3}}}{\theta }\cos \theta d\theta }$

= ${\displaystyle {\int }_{{\sin }^{-1} \frac{1}{3}}^{\frac{\pi }{2}}\frac{{(\sin \theta \times {\cos }^2\theta )}^{\frac{1}{3}}}{{\sin }^4\theta }\cos \theta d\theta }$

= ${\displaystyle {\int }_{{\sin }^{-1} \frac{1}{3}}^{\frac{\pi }{2}}\frac{{\sin }^{\frac{1}{3}}\theta \times {\cos }^{\frac{2}{3}}\theta \cos \theta }{{\sin }^4\theta }d\theta }$

= ${\displaystyle {\int }_{{\sin }^{-1} \frac{1}{3}}^{\frac{\pi }{2}}\frac{{\sin }^{\frac{1}{3}}\theta \cdot {\cos }^{\frac{5}{3}}\theta }{{\sin }^2\theta {\sin }^2\theta }d\theta }$

= ${\displaystyle {\int }_{{\sin }^{-1} \frac{1}{3}}^{\frac{\pi }{2}}\frac{{\cos }^{\frac{5}{3}}\theta }{{\sin }^{\frac{5}{3}}\theta } {\text{cosec}}^2\theta d\theta }$

= ${\displaystyle {\int }_{{\sin }^{-1} \frac{1}{3}}^{\frac{\pi }{2}}{\cot }^{\frac{5}{3}}\theta {\text{cosec}}^2\theta d\theta }$

Put ${\displaystyle \cot \theta =t,-{\text{cosec}}^2\theta d\theta =dt}$

When ${\displaystyle \theta ={\sin }^{-1} \frac{1}{3}}$ or ${\displaystyle \sin \theta =\frac{1}{3},\cot \theta =2\sqrt{2}}$

 ∴ ${\displaystyle t=2\sqrt{2}=\sqrt{8}}$, When ${\displaystyle \theta =\frac{\pi }{2},\cot \theta =0}$

I = ${\displaystyle {\int }_{\sqrt{8}}^0{t}^{\frac{5}{3}}dt={\int }_0^{\sqrt{8}}{t}^{\frac{5}{3}}dt}$

= ${\displaystyle \frac{3}{8}{\left(t^{\frac{8}{3}}\right)}_0^{\sqrt{8}}=\frac{3}{8}{\left(\sqrt{8}\right)}^{\frac{8}{3}}}$

= ${\displaystyle \frac{3}{8}{\left(8\right)}^{\frac{8}{6}}=\frac{3}{8}\cdot {\left(8\right)}^{\frac{4}{3}}=\frac{3}{8}\times 16}$ = 6