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

The Correct Option is A

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.