1. ## Integration by parts

Integrate (x^3 dx) / square root of (1-x^2)

i have to integrate it by parts. i already tried it but my answer waa 0=0 is that possible?

2. ## how?

Originally Posted by confusedgirl
Integrate (x^3 dx) / square root of (1-x^2)

i have to integrate it by parts. i already tried it but my answer waa 0=0 is that possible?
Curious how did you end up with 0?

3. let u = x^3
du = 3x^2 dx0
dv = dx / sq rt of (1-x^2)
v= arcsin x
= x^3 arcsin x - integral of 3x^2 arcsinx dx
u = arcsin x
du = dx / sq rt of (1-x^2)
dv = 3x^2 dx
v = x^3
= x^3 arcsin x - x^3 arcsin x + integral of x^3 dx / sq rt (1-x^2)
0=0

4. first find integrate:sin^-1 (x)
then go on integrating keeping x^3 as the part to differentiate everytime.

5. first find integrate:sin^-1 (x)
then go on integrating keeping x^3 as the part to differentiate everytime.

6. Hello, confusedgirl!

Watch very carefully . . .

$I \;=\;\int\frac{x^3\,dx}{\sqrt{1-x^2}}$

We have: . $I \;=\;\int(x^2)\left[x(1-x^2)^{-\frac{1}{2}}dx\right]$

$\text{By parts: }\;\begin{array}{ccccccc}u &=& x^2 && dv &=& x(1-x^2)^{-\frac{1}{2}}dx \\
du &=& 2x\,dx && v &=& -(1-x^2)^{\frac{1}{2}} \end{array}$

Then: . $I \;=\;-x^2(1-x^2)^{\frac{1}{2}} + \int 2x(1-x^2)^{\frac{1}{2}}dx$

. . For the second integral, use "normal" substitution.

. . We have: . $J \;=\;\int (1-x^2)^{\frac{1}{2}}(2x\,dx)$

. . Let: . $u \,=\,1-x^2\quad\Rightarrow\quad du \,=\,-2x\,dx \quad\Rightarrow\quad 2x\,dx \,=\,-du$

. . Substitute: . $J \;=\;\int u^{\frac{1}{2}}(-du) \;=\;-\int u^{\frac{1}{2}}\,du \:=\:-\tfrac{2}{3}u^{\frac{3}{2}} + C \;=\;-\tfrac{2}{3}(1-x^2)^{\frac{3}{2}} + C$

Therefore: . $I \;=\;-x^2(1-x^2)^{\frac{1}{2}} - \tfrac{2}{3}(1-x^2)^{\frac{3}{2}} + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

. . is often simplified beyond all recognition.

Here's how they do it . . .

We have: . $-x^2(1-x^2)^{\frac{1}{2}} - \tfrac{2}{3}(1-x^2)^{\frac{3}{2}}$

Factor: . $-\tfrac{1}{3}(1-x^2)^{\frac{1}{2}}\bigg[3x^2 + 2(1-x^2)\bigg]$ . . . .
hope you followed that!

. . . . $=\;-\tfrac{1}{3}(1-x^2)^{\frac{1}{2}}\bigg[3x^2 + 2 - 2x^2\bigg]$

. . . . $=\;-\tfrac{1}{3}(1-x^2)^{\frac{1}{2}}(x^2+2)$ . . . . ta-DAA!

7. hi confusedgirl,

$\int{\frac{x^3}{\sqrt{1-x^2}}}dx$

$y=1-x^2\ \Rightarrow\ x^2=1-y$

$\frac{dy}{dx}=-2x\ \Rightarrow\ dy=-2xdx\ \Rightarrow\ x^3dx=-\frac{1}{2}x^2dy=\frac{1}{2}(y-1)dy$

$\frac{1}{2}\int{\frac{(y-1)}{\sqrt{y}}}dy=\frac{1}{2}\int{\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)}dy=\frac{1}{2}\int{\left (y^{\frac{1}{2}}-y^{-\frac{1}{2}}\right)}dy$

This can then be integrated using the power rule.

To integrate by parts...

$\frac{1}{2}\int{\frac{(y-1)}{\sqrt{y}}}dy$

$u=y-1\ du=dy$

$dv=y^{-\frac{1}{2}}\ \Rightarrow\ v=\int{y^{-\frac{1}{2}}}dy=2\sqrt{y}$

$\frac{1}{2}\int{u}dv=\frac{1}{2}\left(uv-\int{v}du\right)=\frac{1}{2}\left((y-1)2\sqrt{y}-\int{2\sqrt{y}}dy\right)$

$\frac{1}{2}\left(-x^2(2)\sqrt{1-x^2}-\frac{4}{3}\left(1-x^2\right)\sqrt{1-x^2}\right)+C$

$=-x^2\sqrt{1-x^2}-\frac{2}{3}\left(1-x^2\right)\sqrt{1-x^2}+C=\sqrt{1-x^2}\left(-x^2-\frac{2}{3}+\frac{2}{3}x^2\right)+C$

$=-\frac{1}{3}\sqrt{1-x^2}\left(x^2+2\right)+C$

8. thanks