What is $\int \frac{4 x}{\sqrt{4 - x^{4}}} d x$ $int (4x ) / sqrt(4-x^4) dx$ ?

Question

1645 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-28T19:21:04

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

Here,

$I = \int \frac{4 x}{\sqrt{4 - x^{4}}} d x$ $I=int(4x)/sqrt(4-x^4)dx$

$x^{2} = 2 sin u \Rightarrow 2 x d x = 2 cos u d u \Rightarrow 4 x d x = 4 cos u d u$ $x^2=2sinu=>2xdx=2cosudu=>4xdx=4cosudu$

$and sin u = \frac{x^{2}}{2} \Rightarrow u = {sin}^{- 1} (\frac{x^{2}}{2})$ $andsinu=x^2/2=>color(blue)(u=sin^-1(x^2/2)$

So,

$I = \int \frac{4 cos u}{\sqrt{4 - 4 {sin}^{2} u}} d u$ $I=int(4cosu)/sqrt(4-4sin^2u)du$

$= \int \frac{4 cos u}{\sqrt{4 {cos}^{2} u}} d u$ $=int(4cosu)/sqrt(4cos^2u)du$

$= \int \frac{4 cos u}{2 cos u} d u$ $=int(4cosu)/(2cosu)du$

$= \int 2 d u$ $=int2du$

$=2u+c...to where,color(blue)(u=sin^-1(x^2/2)$

Hence,

$I=2sin^-1(x^2/2)+c$

What is int (4x ) / sqrt(4-x^4) dx∫4x√4−x4dxint (4x ) / sqrt(4-x^4) dx?