11@ Nishant sir, can we use Walli's formula ??????
This is another question from Larson's book.. .(Today I have given 2-3 from it)
I_n=\int_{0}^{\pi/2}{\sin^nx dx}
Show that
I_{2n}=\frac{1.3.5....(2n-1)}{2.4.6....2n}\times \frac{\pi}{2}
and
I_{2n+1}=\frac{2.4.6....(2n-2)}{1.3.5....(2n-1)}
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7 Answers
3actually dude this was just reduction.And the value of the integral remains the same if we have cosx in place of sinx,
11Yes msp I know walli's formula is just a reduction . But we can prove it easily by walli's formula
3actually i dunno walli,but i know reduction,actually we are discussing this problem two or three days back,do u remember but nishant sir is expecting a different proof not like a normal reduction.
11WALLI'S FORMULA
If m be a positive integer, then
=
(m-1) (m-3) (m-5) .........5.3.1m(m-2)(m-4)....6.4.2(∩/2) , if m is even
and similarily
(m-1) (m-3) (m-5) ........6.4.2m (m-2) (m-4) ....5.3.1 ; if m is odd
24No one trying it ????
In=∫sinnx dx=∫sinx.sinn-1x dx
integrating by parts ..and applying limits 0 to pi/2
In=(n-1)In-2-(n-1)In
=>In=n-1nIn-2
Io=pi/2
I1=1
when n=2k
In=1.3.5.......(2k-1)2.4.6.....2k π2
when n=2k+1
In=2.4.6.....2k1.3.5.......(2k+1) π2
now anyone trying ot prove wallis formula ?????