How do you find the integral $\int x \cdot \sqrt{1 - x^{4}} d x$ $intx*sqrt(1-x^4)dx$ ?

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How do you find the integral $\int x \cdot \sqrt{1 - x^{4}} d x$ $intx*sqrt(1-x^4)dx$ ?

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Answer 1 · 2014-07-27T01:37:50

Answer, $= \frac{1}{4} ({sin}^{- 1} (x^{2}) + x^{2} \sqrt{1 - x^{4}}) + c$ $=1/4(sin^(-1)(x^2)+x^2sqrt(1-x^4))+c$

Explanation :
$\int x \cdot \sqrt{1 - x^{4}} d x = \int x \cdot \sqrt{1 - {(x^{2})}^{2}} d x$ $intx*sqrt(1-x^4)dx = intx*sqrt(1-(x^2)^2)dx$

Using Trigonometric Substitution
let's $x^{2} = sin t, \Rightarrow 2 x d x = cos t d t$ $x^2=sint, => 2xdx=costdt$ . Here I am using $sin t$ $sint$ , it can also be done by considering $x^{2} = cos t$ $x^2=cost$ .

Now plugging in integral,

$= \int \frac{1}{2} \sqrt{1 - x^{4}} (2 x d x)$ $=int1/2sqrt(1-x^4)(2xdx)$

$= \int \frac{1}{2} \cdot cos t \cdot \sqrt{1 - {sin}^{2} t} d t$ $=int1/2*cost*sqrt(1-sin^2t)dt$

Using the identity, ${sin}^{2} t + {cos}^{2} t = 1$ $sin^2t+cos^2t=1$
we get,

$= \int \frac{1}{2} \cdot {cos}^{2} t d t$ $=int1/2*cos^2tdt$ ,

Using another important identity, $cos 2 t = {cos}^{2} t - {sin}^{2} t$ $cos2t=cos^2t-sin^2t$
$\Rightarrow cos 2 t = 2 {cos}^{2} t - 1$ $=>cos2t=2cos^2t-1$

$\Rightarrow {cos}^{2} t = \frac{1 + cos 2 t}{2}$ $=>cos^2t=(1+cos2t)/2$

$= \int \frac{1}{2} \cdot \frac{1}{2} \cdot (1 + cos 2 t) d t = \frac{1}{4} \int 1 d t + \frac{1}{4} \int cos 2 t d t$ $=int1/2*1/2*(1+cos2t)dt = 1/4int1dt+1/4intcos2tdt$

Using Trigonometric functions,

$= \frac{1}{4} \cdot t + \frac{1}{4} \frac{sin 2 t}{2} + c$ $=1/4*t+1/4(sin2t)/2+c$ , where c is constant

$= \frac{1}{4} \cdot (t + \frac{sin 2 t}{2}) + c$ $=1/4*(t+(sin2t)/2)+c$

Use the identity $sin 2 t = 2 sin t cos t$ $sin2t=2sintcost$ :

$= \frac{1}{4} (t + \frac{2 sin t cos t}{2}) + c$ $=1/4(t+(2sintcost)/2)+c$
$= \frac{1}{4} (t + sin t cos t) + c$ $=1/4(t+sintcost)+c$
$= \frac{1}{4} (t + sin t \sqrt{1 - {sin}^{2} t}) + c$ $=1/4(t+sint sqrt(1-sin^2t))+c$

And substituting $sin t = x^{2}$ $sint=x^2$ back in:
$= \frac{1}{4} ({sin}^{- 1} (x^{2}) + x^{2} \sqrt{1 - x^{4}}) + c$ $=1/4(sin^(-1)(x^2)+x^2sqrt(1-x^4))+c$

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How do you find the integral $\int x \cdot \sqrt{1 - x^{4}} d x$ $intx*sqrt(1-x^4)dx$ ?

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