Question

Integration/Reduction formula help. Question attached below. Will give medal

Answer 1

If \[I _{n}= \int\limits_{0}^{\frac{ \pi }{ 2 }}\cos ^{n}dx, \] prove that \[nI _{n}= (n-1)I _{n}-2, (n>1). \].
Evaluate: a) \[\int\limits\limits_{0}^{\frac{ \pi }{ 2 }}\cos ^{6}dx\]
\[b) \int\limits\limits_{0}^{\frac{ \pi }{ 2 }}\cos ^{7}dx\]

Answer 2

maybe we can try induction
assuming n is integer >1

Answer 3

which it is

Answer 4

ok

Answer 5

n=2 would be the base case...
But I don't think both sides of that equality are the same

Answer 6

unless I'm doing something wrong

Answer 7

oh do you mean \[nI_n=(n-1)I_{n-2} ?\]

Answer 8

brb

Answer 9

Yes, that's what I meant

Answer 10

So for n=2; we have
\[\text{ we want to show } 2I_2=(2-1)I_{2-2} \]
\[\text{ so left hand side }=2I_2=2 \int\limits_{0}^{\frac{\pi}{2}}\cos^2(x) dx =\int\limits_{0}^{\frac{\pi}{2}}(1+\cos(2x)) dx =? \\ \text{ I will let you finish that line } \\ \text{ and right hand side } =(2-1)I_0=1I_0=I_0=\int\limits_{0}^{\frac{\pi}{2}}dx=\frac{\pi}{2}\]

Answer 11

Induction Hypothesis:
\[\text{ suppose } kI_k=(k-1)I_{k-2} \text{ for some integer } k>2\]

Answer 12

Now we want to use that to show that:
\[(k+1)I_{k+1}=kI_{k-1}\]

Answer 13

hmm