Question

The value of e integral $∫^1_\frac{1}{3}\frac{(x-x^3)}{x^4}^\frac{1}{3} dx$ is.

• A. 6
• B. 0
• C. 3
• D. 4

Answer 1

Answer: (A)

6

Answer 2

The correct option is(A): 6.

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

$Also,let x=sinθ⇒dx=cosθdθ$

$When x=\frac{1}{3}θ=sin^-1(\frac{1}{3}) and when x=1,θ=\frac{π}{2}$

$⇒I=∫^{π}{2}_sin-1(\frac{1}{3})\frac{(sinθ-sin3θ)^\frac{1}{3}}{sin4θ}cosθdθ$

$=∫^{π}{2}_sin-1(\frac{1}{3})\frac{(sinθ)^\frac{1}{3}(1-sin2θ)^\frac{1}{3}}{sin^4θ}cosθdθ$

$=∫^\frac{π}{2}_sin-1(\frac{1}{3})\frac{(sinθ)^\frac{1}{3}(cosθ)^\frac{2}{3}}{sin^4θ}cosθdθ$

$=∫^{π}{2}_sin-1(^\frac{1}{3})\frac{(sinθ)^\frac{1}{3}(cosθ)^\frac{2}{3}}{sin^2θsin^2θ}cosθdθ$

$=∫^{π}{2}_sin-1(\frac{1}{3})\frac{(cosθ)^\frac{5}{3}}{sinθ)^\frac{5}{3}}cosec^2θdθ$

$=∫^{π}{2}_sin-1(\frac{1}{3}){(cotθ)^\frac{5}{3}}cosec^2θdθ$

$Let cotθ=t -cosec2θdθ=dt$

$When θ=sin-1\frac{1}{3},t=2√2 and when θ=\frac{π}{2},t=0$

$∴I=-∫^0_2√2(t)^\frac{5}{3}dt$

$=-[\frac{3}{8}(t)^\frac{8}{3}]^0_2√2$

$=-\frac{3}{8}[(t)^\frac{8}{3}]^0_√2$

$=-\frac{3}{8}[-(2√2)^\frac{8}{3}]$

$=-\frac{3}{8}[-(√8)^\frac{8}{3}]$

$=\frac{3}{8}[(8)^\frac{4}{3}]$

$=\frac{3}{8}[16]$

$=3×2$

$=6$

Hence,the correct Answer is A.