How do I integrate this ?

J

Jaskaran

Junior Member
Joined
May 5, 2006
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67
Integral of Sin^5(x)Cos^18(x)dx from 0 to pi/2

I'm doing it the reduction method, and come to

[Sin^2(x)]^2*Cos^18(x)*Sin(x) dx

[(1-Cos^2(x)]^2*Cos^18(x)*Sin(x) dx

Using u-substitution, u = cos(x), du = -sin(x)

-?[(1-u^2)^2*u^18]du

Reduces to something like,

-u^72 + 2u^36 - u^18

Plugging cosine of pi/2 would give me 0, right?

But why is the answer according to the online solution listed as 0.000871744578838402?

Yes mad.
 
Re: How do I integrate this *****?

You irate?

Check your polynomial multiplication again. Your highest power after integrating should be 23

When you plug in \(\displaystyle \frac{\pi}{2}\), yes, all cosines will vanish, but when you plug in zero, they won't.
 
\(\displaystyle \int_{0}^{\pi/2}sin(5x)cos(18x) \ dx\)

\(\displaystyle Product-to-Sum \ Formulas: \ sin(5x)cos(18x) \ = \ \frac{1}{2}[sin(23x)+sin(-13x)]\)

\(\displaystyle Ergo, \ we \ have \ \frac{1}{2}\int_{0}^{\pi/2}sin(23x)dx-\frac{1}{2}\int_{0}^{\pi/2}sin(13x)dx\)

\(\displaystyle I'm \ assuming \ that \ you \ can \ take \ it \ from \ here.\)
 
Although the Product-to-Sum identity isn't that often used, these identities were not just "made up" by some bored mathematician. When appropiate, used them and it will save you a lot of grunt work and angst.

After thought: The answer in your on line solution manual is incorrect, the correct answer is -5/299.
 
Glenn, I believe the poster meant

\(\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{5}(x)cos^{18}(x)dx\)

Let \(\displaystyle u=cos(x), \;\ -du=sin(x)dx\)

\(\displaystyle \int(1-cos^{2}(x))^{2}sin(x)cos^{18}(x)dx\)

Making the subs gives:

\(\displaystyle -\int(1-u^{2})^{2}u^{18}du\)

Which indeed results in \(\displaystyle \frac{8}{9177}\approx .0008717446\) when using the integration limits.
 
You're right galactus, I stand corrected.

A good lesson for I, as when everything else fails, read the problem correctly.
 
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