Evaluate: ∫ √(x/4-x3) dx

Question

Evaluate: ∫ √(x/4-x3) dx  (A) (2/3) sin-1 ((x3/2/2)) + c  (B) (2/3) sin-1 (x3/2) + c  (C)  2 sin-1 ((x3/2/2)) + c  (D) (1/3) sin-1 ((x3/2/2)) + c

Tardigrade Admin · Accepted Answer

I = ∫ √(x/4-x3 ) dx = ∫ (√x dx/√4-x3) Here integral of √x = (2/3) and Put x3/2 4 -x3 = 4 - (x3/2)2 x3/2 = t ⇒ √x dx = (2/3) dt So I = (2/3) ∫ (dt/√4-t2 ) = (2/3) sin-1((x3/2/2))+c

Q. Evaluate: $\int \frac{x}{4 - x ^{3}} d x$

$\frac{2}{3} sin^{- 1} (\frac{x ^{3/2}}{2}) + c$

$\frac{2}{3} sin^{- 1} (x^{3/2}) + c$

$2 sin^{- 1} (\frac{x ^{3/2}}{2}) + c$

$\frac{1}{3} sin^{- 1} (\frac{x ^{3/2}}{2}) + c$

$I = \int \frac{x}{4 - x ^{3}} d x = \int \frac{x d x}{4 - x ^{3}}$
Here integral of $x = \frac{2}{3}$ and
Put $x^{3/2} 4 - x^{3} = 4 - (x^{3/2})^{2} x^{3/2} = t$
$\Rightarrow x d x = \frac{2}{3} d t$
So $I = \frac{2}{3} \int \frac{d t}{4 - t ^{2}}$
$= \frac{2}{3} sin^{- 1} (\frac{x ^{3/2}}{2}) + c$

Q. Evaluate: ∫4−x3x​​dx

32​sin−1(2x3/2​)+c

32​sin−1(x3/2)+c

2sin−1(2x3/2​)+c

31​sin−1(2x3/2​)+c

I=∫4−x3x​​dx=∫4−x3​x​dx​ Here integral of x​=32​ and Put x3/24−x3=4−(x3/2)2x3/2=t ⇒x​dx=32​dt So I=32​∫4−t2​dt​ =32​sin−1(2x3/2​)+c

Q. Evaluate: $\int \frac{x}{4 - x ^{3}} d x$

$\frac{2}{3} sin^{- 1} (\frac{x ^{3/2}}{2}) + c$

$\frac{2}{3} sin^{- 1} (x^{3/2}) + c$

$2 sin^{- 1} (\frac{x ^{3/2}}{2}) + c$

$\frac{1}{3} sin^{- 1} (\frac{x ^{3/2}}{2}) + c$