Integration by Parts!

Kimmy2

New member
Joined
Nov 6, 2009
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8
2j4ts3b.png


I need help solving this problem using integration by parts.
I've been struggling with this one.

Thanks so much!w
 
Kimmy2 said:
2j4ts3b.png


I need help solving this problem using integration by parts.
I've been struggling with this one.

Thanks so much!w

u = sin[sup:3g9t9hzb]-1[/sup:3g9t9hzb](4x)

dv = x dx
 
Subhotosh Khan said:
Kimmy2 said:
2j4ts3b.png


I need help solving this problem using integration by parts.
I've been struggling with this one.

Thanks so much!w

u = sin[sup:1mygbsvp]-1[/sup:1mygbsvp](4x)

dv = x dx

Yes, I have gotten that far. I have the integral to be
2fca5e0342eee30528cadeb778192f1.png
 
Kimmy2 said:
Subhotosh Khan said:
Kimmy2 said:
2j4ts3b.png


I need help solving this problem using integration by parts.
I've been struggling with this one.

Thanks so much!w

u = sin[sup:11gta8d5]-1[/sup:11gta8d5](4x)

dv = x dx

Yes, I have gotten that far. I have the integral to be
2fca5e0342eee30528cadeb778192f1.png

Well should have said so in the first place!

Now substitute

2x=sinθ\displaystyle 2x = sin\theta
 
01/4xarcsin(4x)dx, I. by P.      udv = uvvdu.\displaystyle \int_{0}^{1/4}xarcsin(4x)dx, \ I. \ by \ P. \ \implies \ \int udv \ = \ uv-\int vdu.

Let u = arcsin(4x), du = 4[1(4x)2]1/2dx, dv = xdx, v = x22\displaystyle Let \ u \ = \ arcsin(4x), \ du \ = \ \frac{4}{[1-(4x)^{2}]^{1/2}}dx, \ dv \ = \ xdx, \ v \ = \ \frac{x^{2}}{2}

Ergo, 01/4xarcsin(4x)dx = x2arcsin(4x)2]01/4201/4x2[1(4x)2]1/2dx\displaystyle Ergo, \ \int_{0}^{1/4}xarcsin(4x)dx \ = \ \frac{x^{2}arcsin(4x)}{2}\bigg]_{0}^{1/4}-2\int_{0}^{1/4}\frac{x^{2}}{[1-(4x)^{2}]^{1/2}}dx

= π64201/4x2[1(4x)2]1/2dx\displaystyle = \ \frac{\pi}{64}-2\int_{0}^{1/4}\frac{x^{2}}{[1-(4x)^{2}]^{1/2}}dx

Now, let x = sin(θ)4, dx = cos(θ)dθ4, x2 = sin2(θ)16\displaystyle Now, \ let \ x \ = \ \frac{sin(\theta)}{4}, \ dx \ = \ \frac{cos(\theta)d\theta}{4}, \ x^{2} \ = \ \frac{sin^{2}(\theta)}{16}

Hence, = π641320π/2sin2(θ)cos(θ)[1sin2(θ)]1/2dθ\displaystyle Hence, \ = \ \frac{\pi}{64}-\frac{1}{32} \int_{0}^{\pi/2}\frac{sin^{2}(\theta)cos(\theta)}{[1-sin^{2}(\theta)]^{1/2}}d\theta

= π641320π/2sin2(θ)cos(θ)cos(θ)dθ\displaystyle = \ \frac{\pi}{64}-\frac{1}{32}\int_{0}^{\pi/2}\frac{sin^{2}(\theta)cos(\theta)}{cos(\theta)}d\theta

= π641320π/2sin2(θ)dθ = π641320π/21cos(2θ)2dθ\displaystyle = \ \frac{\pi}{64}-\frac{1}{32}\int_{0}^{\pi/2}sin^{2}(\theta)d\theta \ = \ \frac{\pi}{64}-\frac{1}{32}\int_{0}^{\pi/2}\frac{1-cos(2\theta)}{2}d\theta

= π641640π/2[1cos(2θ)]dθ = π64164[θsin(θ)cos(θ)]0π/2\displaystyle = \ \frac{\pi}{64}-\frac{1}{64}\int_{0}^{\pi/2}[1-cos(2\theta)]d\theta \ = \ \frac{\pi}{64}-\frac{1}{64}[\theta-sin(\theta)cos(\theta)\bigg]_{0}^{\pi/2}

= π64164[π20(00)] = π64π128 = π128.\displaystyle = \ \frac{\pi}{64}-\frac{1}{64}[\frac{\pi}{2}-0-(0-0)] \ = \ \frac{\pi}{64}-\frac{\pi}{128} \ = \ \frac{\pi}{128}.

Therefore, 01/4xarcsin(4x)dx = π128, QED\displaystyle Therefore, \ \int_{0}^{1/4}xarcsin(4x)dx \ = \ \frac{\pi}{128}, \ QED
 
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