integration by parts and trig substitution

[thebenji]

The problem:

. . .int[x=0 to 1/3] [x sin^-1(3x)] dx

Here is what I am doing to reach the incorrect answer:

. . .u = (sin(3x))^-1

. . .dv = x dx

. . .du = ( 3 / sqrt[1 - 9x²] ) dx

. . .v = (x²) / 2

. . .int[x=0 to 1/3] [x sin^-1(3x)] dx

. . . . .= (1/2) x² sin^-1(3x) - (1/2) int [ (3x²) / sqrt[1 - 9x²] ] dx


u = x^2. . . . .dv = 3/[sqrt(1 - 9x^2)] dx
du = 2x dx. . . . .v = arcsin(3x)

=(1/2)(x^2)(arcsin(3x)) - (x^2)(arcsin(3x)) + 2*int[(x)(arcsin(3x))dx]

-int[x*(arcsin(3x))]dx = (1/2)(x^2)(arcsin(3x)) - (x^2)(arcsin(3x))

int[x*(arcsin(3x))]dx = (x^2)(arcsin(3x)) - (1/2)(x^2)(arcsin(3x))

Evaluate both parts of right hand side from 0 to (1/3):

= [(1/9)(pi/2) - 0] - (1/2)[(1/9)(pi/2)]

= (pi/18) - (1/18)(pi/2)

= (pi/18) - (pi/36)

= (2pi/36) - (pi/36)

= (pi/36)

What am I doing wrong ---- how do I solve this problem?
___________________
Edited by stapel -- Reason for edit: removing hotlinks

 

[soroban]

Hello, thebenji!

Your work is correct up to the second stage . . .

\(\displaystyle \L\;\;\int^{\;\;\;\frac{1}{3}}_0 x\,\arcsin(3x)\,dx\)

You integrated by parts and got:

\(\displaystyle \L\;\;\frac{1}{2}x^2\,\arcsin(3x) - \frac{3}{2}\int\frac{x^2}{\sqrt{1\,-\,9x^2}}\,dx\;\;\) . . . Correct!


At this point, I'd use a Trig Substitution.

Let \(\displaystyle 9x^2\,=\,\sin^2\theta\;\;\Rightarrow\;\;x\,=\,\frac{1}{3}\sin\theta\;\;\Rightarrow\;\;dx \,=\,\frac{1}{3}\cos\theta\,d\theta\)
\(\displaystyle \;\;\) and: \(\displaystyle \sqrt{1\,-\,9x^2} \:=\:\sqrt{1\,-\,\sin^2\theta} \:=\:\sqrt{\cos^2\theta} \:=\:\cos\theta\)

The second integral becomes:
\(\displaystyle \L\;\;\int\frac{x^2}{\sqrt{1\,-\,9x^2}}\,dx\;=\;\int\frac{\frac{1}{9}\sin^2\theta}{\cos\theta}\left(\frac{1}{3}\cos\theta\,d\theta\right) \;=\;\frac{1}{27}\int\sin^2\theta\,d\theta\)

Can you finish it now?