Integration by Parts!
- Thread starter Kimmy2
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D
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Kimmy2 said:
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:
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
D
Deleted member 4993
Guest
Kimmy2 said:Subhotosh Khan said:Kimmy2 said:
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
Well should have said so in the first place!
Now substitute
\(\displaystyle 2x = sin\theta\)
BigGlenntheHeavy
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\(\displaystyle \int_{0}^{1/4}xarcsin(4x)dx, \ I. \ by \ P. \ \implies \ \int udv \ = \ uv-\int vdu.\)
\(\displaystyle Let \ u \ = \ arcsin(4x), \ du \ = \ \frac{4}{[1-(4x)^{2}]^{1/2}}dx, \ dv \ = \ xdx, \ v \ = \ \frac{x^{2}}{2}\)
\(\displaystyle Ergo, \ \int_{0}^{1/4}xarcsin(4x)dx \ = \ \frac{x^{2}arcsin(4x)}{2}\bigg]_{0}^{1/4}-2\int_{0}^{1/4}\frac{x^{2}}{[1-(4x)^{2}]^{1/2}}dx\)
\(\displaystyle = \ \frac{\pi}{64}-2\int_{0}^{1/4}\frac{x^{2}}{[1-(4x)^{2}]^{1/2}}dx\)
\(\displaystyle Now, \ let \ x \ = \ \frac{sin(\theta)}{4}, \ dx \ = \ \frac{cos(\theta)d\theta}{4}, \ x^{2} \ = \ \frac{sin^{2}(\theta)}{16}\)
\(\displaystyle Hence, \ = \ \frac{\pi}{64}-\frac{1}{32} \int_{0}^{\pi/2}\frac{sin^{2}(\theta)cos(\theta)}{[1-sin^{2}(\theta)]^{1/2}}d\theta\)
\(\displaystyle = \ \frac{\pi}{64}-\frac{1}{32}\int_{0}^{\pi/2}\frac{sin^{2}(\theta)cos(\theta)}{cos(\theta)}d\theta\)
\(\displaystyle = \ \frac{\pi}{64}-\frac{1}{32}\int_{0}^{\pi/2}sin^{2}(\theta)d\theta \ = \ \frac{\pi}{64}-\frac{1}{32}\int_{0}^{\pi/2}\frac{1-cos(2\theta)}{2}d\theta\)
\(\displaystyle = \ \frac{\pi}{64}-\frac{1}{64}\int_{0}^{\pi/2}[1-cos(2\theta)]d\theta \ = \ \frac{\pi}{64}-\frac{1}{64}[\theta-sin(\theta)cos(\theta)\bigg]_{0}^{\pi/2}\)
\(\displaystyle = \ \frac{\pi}{64}-\frac{1}{64}[\frac{\pi}{2}-0-(0-0)] \ = \ \frac{\pi}{64}-\frac{\pi}{128} \ = \ \frac{\pi}{128}.\)
\(\displaystyle Therefore, \ \int_{0}^{1/4}xarcsin(4x)dx \ = \ \frac{\pi}{128}, \ QED\)
\(\displaystyle Let \ u \ = \ arcsin(4x), \ du \ = \ \frac{4}{[1-(4x)^{2}]^{1/2}}dx, \ dv \ = \ xdx, \ v \ = \ \frac{x^{2}}{2}\)
\(\displaystyle Ergo, \ \int_{0}^{1/4}xarcsin(4x)dx \ = \ \frac{x^{2}arcsin(4x)}{2}\bigg]_{0}^{1/4}-2\int_{0}^{1/4}\frac{x^{2}}{[1-(4x)^{2}]^{1/2}}dx\)
\(\displaystyle = \ \frac{\pi}{64}-2\int_{0}^{1/4}\frac{x^{2}}{[1-(4x)^{2}]^{1/2}}dx\)
\(\displaystyle Now, \ let \ x \ = \ \frac{sin(\theta)}{4}, \ dx \ = \ \frac{cos(\theta)d\theta}{4}, \ x^{2} \ = \ \frac{sin^{2}(\theta)}{16}\)
\(\displaystyle Hence, \ = \ \frac{\pi}{64}-\frac{1}{32} \int_{0}^{\pi/2}\frac{sin^{2}(\theta)cos(\theta)}{[1-sin^{2}(\theta)]^{1/2}}d\theta\)
\(\displaystyle = \ \frac{\pi}{64}-\frac{1}{32}\int_{0}^{\pi/2}\frac{sin^{2}(\theta)cos(\theta)}{cos(\theta)}d\theta\)
\(\displaystyle = \ \frac{\pi}{64}-\frac{1}{32}\int_{0}^{\pi/2}sin^{2}(\theta)d\theta \ = \ \frac{\pi}{64}-\frac{1}{32}\int_{0}^{\pi/2}\frac{1-cos(2\theta)}{2}d\theta\)
\(\displaystyle = \ \frac{\pi}{64}-\frac{1}{64}\int_{0}^{\pi/2}[1-cos(2\theta)]d\theta \ = \ \frac{\pi}{64}-\frac{1}{64}[\theta-sin(\theta)cos(\theta)\bigg]_{0}^{\pi/2}\)
\(\displaystyle = \ \frac{\pi}{64}-\frac{1}{64}[\frac{\pi}{2}-0-(0-0)] \ = \ \frac{\pi}{64}-\frac{\pi}{128} \ = \ \frac{\pi}{128}.\)
\(\displaystyle Therefore, \ \int_{0}^{1/4}xarcsin(4x)dx \ = \ \frac{\pi}{128}, \ QED\)