How do you integrate $\int \frac{x}{\sqrt{16 - x^{4}}} d x$ $int x /sqrt( 16 - x^4 )dx$ using trigonometric substitution?

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Answer 1 · 2016-01-21T18:07:46

you just need to do a normal substitution

$u = \frac{1}{4} x^{2}$ $u = 1/4x^2$
$d u = \frac{1}{2} x d x$ $du = 1/2xdx$

$2 \int \frac{1}{2} \frac{x}{\sqrt{16 - x^{4}}} d x$ $2int1/2x/sqrt(16-x^4)dx$

$2 \int \frac{1}{\sqrt{16 - 16 u^{2}}} d u$ $2int1/sqrt(16-16u^2)du$

$\frac{1}{2} \int \frac{1}{\sqrt{1 - u^{2}}} d u$ $1/2int1/sqrt(1-u^2)du$

as you know, it's the derivate of $arcsin (u)$ $arcsin(u)$

so

$= \frac{1}{2} [arcsin (u)] + C$ $= 1/2[arcsin(u)]+C$

$= \frac{1}{2} [arcsin (x^{2})] + C$ $= 1/2[arcsin(x^2)]+C$

OF COURSE You can do it with $u = sin (t)$ $u = sin(t)$

$d u = cos (t) d t$ $du = cos(t) dt$

and you have

$\frac{1}{2} \int \frac{cos (t)}{cos (t)} d t$ $1/2intcos(t)/cos(t)dt$

$\frac{1}{2} \int d t$ $1/2intdt$

$\frac{1}{2} [t]$ $1/2[t]$

the fact is $t = arcsin (u)$ $t = arcsin(u)$ ...

How do you integrate int x /sqrt( 16 - x^4 )dx∫x√16−x4dxint x /sqrt( 16 - x^4 )dx using trigonometric substitution?