How do I integrate this ?

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 π2\displaystyle \frac{\pi}{2}, yes, all cosines will vanish, but when you plug in zero, they won't.
 
0π/2sin(5x)cos(18x) dx\displaystyle \int_{0}^{\pi/2}sin(5x)cos(18x) \ dx

ProducttoSum Formulas: sin(5x)cos(18x) = 12[sin(23x)+sin(13x)]\displaystyle Product-to-Sum \ Formulas: \ sin(5x)cos(18x) \ = \ \frac{1}{2}[sin(23x)+sin(-13x)]

Ergo, we have 120π/2sin(23x)dx120π/2sin(13x)dx\displaystyle Ergo, \ we \ have \ \frac{1}{2}\int_{0}^{\pi/2}sin(23x)dx-\frac{1}{2}\int_{0}^{\pi/2}sin(13x)dx

Im assuming that you can take it from here.\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

0π2sin5(x)cos18(x)dx\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{5}(x)cos^{18}(x)dx

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

(1cos2(x))2sin(x)cos18(x)dx\displaystyle \int(1-cos^{2}(x))^{2}sin(x)cos^{18}(x)dx

Making the subs gives:

(1u2)2u18du\displaystyle -\int(1-u^{2})^{2}u^{18}du

Which indeed results in 89177.0008717446\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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