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How???????????????????????????????

I(n) = integral of (sin x)^n

using substitution (sin x)^n=sin x (sin x)^n-1

integrate and solve by parts

u need to use substitution later that (cos x)^2 = 1- (sin x)^2

answer is I(n) = -(1/n) cos x (sin x)^n-1 + ((n-1)/n) I(n-2)

i get it down to the bit where u make the cos^2 substitution but after that i am lost please help me
latentcorpse
I(n) = integral of (sin x)^n

using substitution (sin x)^n=sin x (sin x)^n-1



By parts

I(n) = -cosx(sinn-1x) - int ( - cosx . (n-1)sinn-2xcosx

I(n) = -cosx(sinn-1x) + int (1-sin2x)((n-1)sinn-2x)

I(n) = -cosx(sinn-1x) + (n-1)int(sinn-2x) - (n-1)int(sinnx)

I(n) = -cosx(sinn-1x) + (n-1) I(n-2) - (n-1) I(n)

nI(n) = -cosx(sinn-1x) + (n-1) I(n-2)

I(n) = -cosx(sinn-1x)/n + (n-1)/n I(n-2)
I[sinnx dx] = -cosx sinn-1x + I[cosx (n-1) sinn-2x cosx dx]
= -cosx sinn-1x + (n-1)I[cos[syup]2x sinn-2x dx]
= -cosx sinn-1x + (n-1)I[(1-sin2x) sinn-2x dx]
= -cosx sinn-1x + (n-1)I[sinn-2x dx] - (n-1)I[sinnx dx]
= -cosx sinn-1x + (n-1)In-2 - (n-1)In

Therefore

In = -cosx sinn-1x + (n-1)In-2 - (n-1)In
In+(n-1)In = -cosx sinn-1x + (n-1)In-2
nIn = -cosx sinn-1x + (n-1)In-2

In = (-1/n)cosx sinn-1x + [(n-1)/n]In-2

For the love of god, why doesn't Latex work any more?
Reply 3
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latentcorpse
OP
12
cheers i forgot to substitute the I(n-2) for sin^(n-2)

stupid computational error on my behalf

my bad!

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