reduction help (1 Viewer)

[FD3S-R]

reduction formula questions, seems easy but im gettin real close but not the answers

both Cambridge 5.5

17. If In = ∫sinⁿx dx for n≥0 show that In = -1/n cosxsinⁿֿ¹x + (n-1)/n In-2 for n≥2





18. If In = ∫secⁿx dx for n≥0, then In = 1/(n-1) tanxsecⁿֿ²x + (n-2)/(n-1) In-2 for n≥2



thanks

 

[KFunk]

18.

∫secⁿx dx
= ∫sec^n-2xsec²x dx

so, using int. by parts where u = sec^n-2x and v'=sec²x
hence v= tanx

u = sec^n-2x = (cosx)^2-n
u' = (n-2)sec^n-1xsinx (I can go over this bit of differentiation if needed)

so that:
∫secⁿx dx
= tanxsec^n-2x - ∫(n-2)sec^n-1xsinxtanx
=tanxsec^n-2x - ∫(n-2)secⁿx(1-cos²x)
=tanxsec^n-2x - (n-2)∫secⁿx +(n-2)∫sec^n-2x
= tanxsec^n-2x - (n-2)I_n +(n-2)I_n-2
[since I_n=∫secⁿx dx]

hence I_n= 1/(n-1) tanxsec^n-2x + (n-2)/(n-1)I_n-2

since I_n+ (n-2)I_n = tanxsec^n-2x +(n-2)I_n-2

 

[shafqat]

reduction formula questions, seems easy but im gettin real close but not the answers

both Cambridge 5.5

17. If In = ∫sinⁿx dx for n≥0 show that In = -1/n cosxsinⁿֿ¹x + (n-1)/n In-2 for n≥2





18. If In = ∫secⁿx dx for n≥0, then In = 1/(n-1) tanxsecⁿֿ²x + (n-2)/(n-1) In-2 for n≥2



thanks

for the first one, split sinⁿx into sinⁿֿ¹x.sinx and then apply integration by parts.
similarly split secⁿx into secⁿֿ²x and sec^2x and do likewise.

wat kfunk just did

 

[FD3S-R]

omg wtf.... i did secⁿˉ²xsec²x and then changed to secⁿˉ²x(1+tan²x)

dammit i spent like half a hour checking algebraic errors thanks guys

sorry last quick question

Cambridge Diagnostic Test 5 q13
∫√(sinx)cos³x dx


thx!!!!

 

[KFunk]

let u=sinx
du=cosxdx

∫√(sinx)cos³xdx

=∫√(u)(1-u²)du

=∫ u^1/2 - u^5/2du

=2/3 sin^3/2x - 2/7sin^7/2x +c

Why does my font keep going small???...ahh, fixed it.

 

[FD3S-R]

for the first one, split sinⁿx into sinⁿֿ¹x.sinx and then apply integration by parts.
similarly split secⁿx into secⁿֿ²x and sec^2x and do likewise.

wat kfunk just did

we cant split first one since it says for n≥2

i cant get it to work with sin n-2 x

 

[shafqat]

it'll work
try int. by parts with v' = sinx and u = sinⁿֿ¹x

 

[KFunk]

I'll write it up briefly, I got a lot more free time now that the exams are over (and this is about all the free time I'm gonna spend on math ).

I_n= ∫sinⁿx dx
= ∫sin^n-1xsinx dx

Then int. by parts where. u = sin^n-1x and v' = sinx
so u'= (n-1)sin^n-2xcosx (found by substituting p=sinx then du/dx = du/dp.dp/dx) and v= -cosx

I_n= -cosxsin^n-1x + (n-1)∫sin^n-2xcos²x dx
= -cosxsin^n-1x +(n-1)∫sin^n-2x -(n-1)∫sinⁿx
= -cosxsin^n-1x +(n-1)I_n-2 - (n-1)I_n

I_n +(n-1)I_n= -cosxsin^n-1x +(n-1)I_n-2

therefore:
I_n = -1/n(cosxsin^n-1x) +(n-1)/n.I_n-2

 

[FD3S-R]

but the questions states n≥2

so we cant use sinxsin^(n-1)x can we?

 

[KFunk]

but the questions states n≥2

so we cant use sinxsin^(n-1)x can we?

I don't see why not for n≥2

if n=2 then
sinxsin^n-1x = sinxsinx i.e it works fine.

You have to split it up like that in order to prove the form that the question is asking for: (-1/n cosxsinⁿֿ¹x + (n-1)/n In-2) if you don't split it into (sinxsin^n-1x) then you wont get (-cosxsinⁿֿ¹x).


EDIT: In fact, I think half the reason that inequality exists is to give the method validity.

 

[FD3S-R]

this is why i hate reduction i cant clear this up

in reduction sometimes they give u values n exist in

for example n≥2

and the equation cos^nx

does that mean i cant use cosxcos^(n-1)x and must result into cos^3xcos^(n-3)x

or cos^2xcos^(n-2)x etc???

 

[KFunk]

They give you the inequality so that you can use the method you need to use. Basically, n must be greater than or equal to 2 otherwise your manipulations would be rendered invalid. Say we're working with:

cosⁿx which we can express as cos^n-1xcosx, if n=1 then:
cos^n-1xcosx = cosx .... in these questions we're often integrating trigonometric functions so note that ∫cosxdx ≠ ∫ 1dx (I'm sure there are some exceptions to this but for the purpose of these types of questions a lot of the processes won't make much sense when n is less than a certain number).

I know I didn't explain that the best way possible, but keep asking if it's still not clear and maybe someone else can reason it out for you.