Question

I'm trying to prove this :
\[\int \sin^n x\cos^m x dx=\frac{-\sin^{n-1}x\cos^{m+1}x}{n+m}+\frac{n-1}{n+m}\int \sin^{n-2}x \cos^mx dx\]

Answer 1

this is avery interesting identities for this kind of integrals

Answer 2

mathematica.stackexchange.com

Better for proofs

Answer 3

i was trying with \[u=\sin^{n-1}x\cos^mx\]
\[dv=\sin x\]

Answer 4

but it was very messy calculations lol

Answer 5

i have also thought of doing something of this sort
\[\sin^nx=\sin^2x\sin^{n-2}x \]

Answer 6

i thought of making it somewhat easy to integrate problem but i hit the wall lol

Answer 7

@jim_thompson5910

Answer 8

the last cos should be power m not n

Answer 9

so you have tried to do integration by parts?

Answer 10

yes that is my attempt

Answer 11

was*

Answer 12

'm lagging did you write something?

Answer 13

\[\sin^{n-1}(x) \sin(x) \cos^m(x)=\sin^n(x)\cos^m(x) \\ \int\limits_{}^{} \sin^{n-1}(x) \cdot \cos^m(x) \sin(x) dx \\ \]
integrate the cos^m(x)*sin(x) and derivative the sin^(n-1)(x)
let's see what happens ...

Answer 14

hmm let me try that
i did the opposite i chose dv=sinx instead of sin^(n-1)x

Answer 15

\[I=\int\limits \sin^{n-1}(x) \cdot \cos^m(x) \sin(x) dx \\ I=\sin^{n-1}(x) \cdot \frac{-\cos^{m+1}(x)}{m+1} - \int\limits (n-1)\sin^{n-2}(x) \cos(x) \cdot \frac{-\cos^{m+1}(x)}{m+1} dx\]
\[I=\frac{-1}{m+1} \sin^{n-1}(x) \cos^{m+1}(x)+\frac{n-1}{m+1}\int \sin^{n-2}(x) \cos^{m+2} dx\]
is this what you got?

Answer 16

for the first round

Answer 17

I'm still in the process i should have something
let me see

Answer 18

I missed the (x) thingy at the end of my line

Answer 19

we are in the same page

Answer 20

don't worry about it i got you:)

Answer 21

I guess we can see what happens if we apply integration by parts again :p

Answer 22

oh i got it

Answer 23

you got it?

Answer 24

i mean i can see it from here :)
i was stuck on using the right u and dv

Answer 25

coolness :)

Answer 26

sometimes i touch around the problem i get close and get shy haha

Answer 27

like here i was on the right road but didn't manage it

Answer 28

because it could mean a lot of work and no happiness :(

Answer 29

yes indeed!

Answer 30

you hit the answer right way that's quite a skill you have:)

Answer 31

away*

Answer 32

i have another intersting integral i want to challenge it and see what can i come up with
\[\int_{0}^{\pi/2}\sin^nxdx= \] don't even remember the rest lol some polynomial

Answer 33

i manged to see that part actually that is why i made sinx sin^(n-1)x cos^mx too

Answer 34

I keep mistyping what I mean so I gave up

Answer 35

polynnomial?

Answer 36

hahaha no worries :)
thanks a lot, will try to tackle that problem later
i have 3 more to do all of them have to do with by parts

Answer 37

yeah i have to prove it can be written as a polynomial or something
i don't remember the rest. i need to look at the notes
i think the problem are taken from calculus sullivan

Answer 38

n is natural right

Answer 39

yes sure
it is about cases n odd n even
they give different results

Answer 40

then part b asks about polynomial something

Answer 41

right
even n gives us irrational number (the irrational part being the pi thing)
and the odd n gives us rational number
I was trying to find a pattern between the answers
(well looking at the evens and odds separate)

Answer 42

yeah it is about that

Answer 43

I want to go ahead and see if we can look at the odd n in general
let n=2k+1 where k>0 is an integer
\[\int\limits_{0}^{\frac{\pi}{2}}\sin^{2k+1}(x) dx \\ \int\limits_{0}^{\frac{\pi}{2}} \sin(x) \sin^{2k}(x) dx \\ \int\limits_0^{\frac{\pi}{2}} \sin(x)(1-\cos^2(x))^k dx\]

Answer 44

here we can do a sub

Answer 45

yes seems u=cosx

Answer 46

\[u=\cos(x) \\ du=-\sin(x) dx \\ -\int\limits_{1}^{0}(1-u^2)^k du \\ \int\limits_0^1(1-u^2)^k du\]

Answer 47

thinking...

Answer 48

hmm good

Answer 49

http://openstudy.com/users/freckles#/updates/54de8a16e4b0b0ad8854cba3
we did something really cool here
that I'm thinking maybe we can do something similar here
maybe

Answer 50

But..
I have bad news

Answer 51

I have to retire tonight
I shall be back tomorrow hopefully

Answer 52

eh i'm leaving too
i will try myself to see if there is anything i can come up with
those triple integral make me dizzy hahaha

Answer 53

I was talking about the result we derived there
\[I(m,n) =\int\limits_0^a x^m(a-x)^n = \dfrac{m!n!}{(m+n+1)!}a^{m+n+1}\]
This result was obtained by integration by parts

Answer 54

oh i see

Answer 55

We will of course not be able to use that result because our integrand is a bit different

Answer 56

but the idea is related to that correct

Answer 57

but we might be able to do integration by parts to obtain some kinda output like that above

Answer 58

oh i see

Answer 59

I could be wrong
I haven't tried it

Answer 60

but i will tomorrow

Answer 61

well it is worth trying anyway :)

Answer 62

\[I=\int\limits_{0}^{1}(1-u^2)^k du \\ I= \int\limits_0^11 \cdot (1-u^2)^k du \\ I_1=u(1-u^2)^k|_0^1-\int\limits_0^1 u k (1-u^2)^{k-1} (-2u) du \\ I_1=\int\limits_0^1 k(1-u^2)^{k-1} 2u^2 du \\ I_2=\frac{2 u^3}{3} k (1-u^2)^{k-1}|_0^1-\int\limits_0^1 \frac{2u^3}{3}k(k-1)(1-u^2)^{k-2}(-2u) du \\ I_2=\int\limits_0^1 \frac{4 u^4}{3}k(k-1)(1-u^2)^{k-2} du \]
We need to do this k times
for I_k we should have (1-u^2)^(k-k) which means we will able to integrate other little factor u^some power .
We can stop when we see the pattern:
\[I_3=\frac{4 u^5}{ 3 \cdot 5 }k(k-1)(1-u^2)^{k-2}|_0^1-\int\limits_0^1 \frac{4 u^5}{3 \cdot 5 } k(k-1)(k-2)(1-u^2)^{k-3}(-2u) du \\ I_3=\int\limits_0^1 \frac{8 u^{6}}{3 \cdot 5} k(k-1)(k-2)(1-u^2)^{k-3} du\]
.... hmm.. so the pattern looks sorta like this:
\[I_n=\int\limits_0^1 \frac{2^n u^{2n}}{1 \cdot 3 \cdot 5 \cdot 7\cdots (2n-1)}[k(k-1)(k-2) \cdots (k-n)](1-u^2)^{k-n} du \]
replace n with k since we wanted k times
\[I_k=\int\limits_0^1 \frac{2^k u^{2k}}{\Pi_{i=1}^{k}(2i-1)} k! (1-u^2)^{k-k} du \\ I_k=\int\limits_0^1 \frac{2 ^k u^{2k}}{\Pi_{i=1}^{k} (2i-1)} k! du =\frac{2^k k!}{\Pi_{i=1}^{k}(2i-1)} \cdot \frac{u^{2k+1}}{2k+1}|_0^1 \\ =\frac{2^k k!}{(2k+1) \cdot \Pi_{i=1}^{k}(2i-1)}\]
We might want to check that and see if I made any error .

Answer 63

so that means we have
\[I_n=\int\limits_{0}^{\pi/2}\sin^nxdx \\ I_{2n+1}=\frac{2^k k!}{(2k+1) \cdot \Pi_{i=1}^{k}(2i-1)}\]

Answer 64

Now we need to think of the even case

Answer 65

and honestly I think about this case a little more

Answer 66

type-o in my one line:
\[I_n=\int\limits_0^1 \frac{2^n u^{2n}}{1 \cdot 3 \cdot 5 \cdot 7\cdots (2n-1)}[k(k-1)(k-2) \cdots (k-(n-1))](1-u^2)^{k-n} du\]

Answer 67

thanks @freckles i was pretty close to that pattern you found too

Answer 68

I can't find a way to get something like that for the even n route

Answer 69

@ganeshie8 Is it possible to get a solution similar to that the even route way ?

Answer 70

@freckles all the definite integrals in this thread are just a special case of

\[I(m,n) =\int\limits_0^a x^m(a-x)^n = \dfrac{m!n!}{(m+n+1)!}a^{m+n+1}\]

Answer 71

\[\int\limits_{0}^{1}(1-u^2)^k du \]
let \(u^2 = t \implies du = \dfrac{t^{-1/2}dt}{2}\)

the integral becomes
\[\frac{1}{2}\int\limits_0^1 t^{-1/2} (1-t)^k dt = \dfrac{(-1/2)! k!}{2(-1/2+k+1)!}= \dfrac{k!}{2(k+1/2)!}\sqrt{\pi}\]


this is probably not what the assignment is about.. this form certainly looks a lot compact however xD

Answer 72

also we dont need to consider cases (n odd, even) if we allow rational and negative factorials in answer..

Answer 73

Let \(n\) be any integer :
\[\int\limits_{0}^{\pi/2}\sin^nxdx= \int\limits_{0}^{\pi/2}(\sin^2 x)^{(n-1)/2} \sin xdx = \frac{1}{2}\int\limits_0^1 u^{-1/2}(1-u)^{(n-1)/2} du \\~\\~\\= \dfrac{(\frac{-1}{2})! (\frac{n-1}{2})!}{2(\frac{-1}{2}+\frac{n-1}{2}+1)!} = \dfrac{ (\frac{n-1}{2})!}{2(\frac{n}{2})!} \sqrt{\pi}\]

Answer 74

what about deriving
\[\int_{0}^{\pi/2}\sin^nxdx=\frac{(n-1)(n-3)......(4)(2)}{n(n-2)...........(5)(3)(1)}\]
for n odd

Answer 75

we use the main formula :
\[\int \sin^n x\cos^m x dx=\frac{-\sin^{n-1}x\cos^{m+1}x}{n+m}+\frac{n-1}{n+m}\int \sin^{n-2}x \cos^mx dx\]

set \(m=0\) and plugin the bounds

Answer 76

@ganeshie8 you mean for my last comment i should use that
to derive that fraction

Answer 77

that will do right ? why derive everything again..

Answer 78

because the first question is to derive that massive fraction