Reduction formula question from my trial (1 Viewer)

Hermes1 · Aug 4, 2011

$I_{n} = \int_{0}^{\pi }\frac{1-\cos nx}{1-\cos x}$

$\int_{0}^{\frac{\pi }{2}}\frac{\sin ^{2}3\theta }{\sin ^{2}\theta } = \frac{3\pi }{2}$

i dont no if u need the first part to do the second. for the second part it said also show....

Hermes1 · Aug 4, 2011

$\int_{0}^{\frac{\pi }{2}}\frac{\sin ^{2}3\theta +\cos ^{2}3\theta-(\cos ^{2}3\theta -\sin ^{2}3\theta ) }{\sin ^{2}\theta +\cos ^{2}\theta -(\cos ^{2}\theta -\sin ^{2}\theta )}d\theta$

the twos on the top cancel out so nothing needs to be taken out the front

then i changed it to:

$\int_{0}^{\frac{\pi }{2}}\frac{1-\cos 6\theta }{1-\cos 2\theta }d\theta$

was this on the right track?

XTsquared · Aug 4, 2011

Yes, then let 2@ = x (@ = theta)

2d@ = dx
@ = 0, x = 0
@ = pi/2 , x = pi

then work it out (it's in form of (1/2)I(3) half I three)

Hermes1 · Aug 4, 2011

XTsquared said:
Yes, then let 2@ = x (@ = theta)

2d@ = dx
@ = 0, x = 0
@ = pi/2 , x = pi

then work it out (it's in form of (1/2)I(3) half I three)

oh crap. i did something stupid in the exam and i said let k = n/2 or sumfin stupid. ur way looks right.

hup · Aug 5, 2011

Could you give the whole question? I want to try it

Hermes1 · Aug 5, 2011

hup said:
Could you give the whole question? I want to try it

when i get my paper back ill post the whole question. i just cant quite remember it fully.

Reduction formula question from my trial (1 Viewer)

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