Sydney Grammar integration question (1 Viewer)

Synchronised · Dec 21, 2012

How do I do this question:
Prove that $\int_{0}^{\frac{\pi }{2}}\frac{cos5\theta sin\theta }{cos\theta }d\theta =\frac{23}{15}$
thanks in advance.

kenkap · Dec 21, 2012

what happens when u let u=cos(theta)??? try that..

Synchronised · Dec 21, 2012

kenkap said:
what happens when u let u=cos(theta)??? try that..

Oh wait, there are 2 parts to this question.
Part 1: If $I_{n}=\int sin n\theta sec\theta d\theta$ find $I_{n}$ which is $I_{n}=\frac{2cos(n-1)\theta }{n-1}-I_{n-2}$ and the second part is hence prove...
so I don't think I can use the substitution.

Synchronised · Dec 21, 2012

Drongoski said:
Use the identity:

2cosA sinB = sin(A+B) - sin(A-B)

Got it, thanks

Drongoski · Dec 21, 2012

No. Please ignore the above identity!

$cos 5 \theta = 16cos^5 \theta - 20cos^3 \theta + 5cos \theta$

$\therefore I = \int _0 ^{\frac {\pi}{2}} (16cos^4\theta -20cos^2\theta+5)dcos\theta$

$= - \left[ \frac {16cos^5 \theta}{5} - \frac {20cos^3 \theta}{3} + 5cos \theta\right]^{\frac{\pi}{2}} _0$

$= -(0 - 0 + 0) + (\frac {16}{5} - \frac {20}{3} + 5) = \frac {23}{15}$

Synchronised · Dec 21, 2012

Drongoski said:
No. Pleasae ignore the identity!

$cos 5 \theta = 16cos^5 \theta - 20cos^3 \theta + 5cos \theta$

but it works fine with the identity, I get: $\frac{1}{2}\int_{0}^{\frac{\pi }{2}}sin6\theta sec\theta -sin4\pi sec\theta d\theta$ which can be evaluated using part a.

Sacredoblivion · Jan 28, 2013

Lolol, I remember doing this

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Sydney Grammar integration question (1 Viewer)

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