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integration FP1

JohnKennedy

Hey yall,

how would you integrate 2sin²xcosx?

Many thanks

Reply 1

Glutamic Acid

I would make a substitution.

Reply 2

Daniel Freedman

What is the relationship between sinx and cosx? Can you see that you have something in the form:

k[f(x)]^n f'(x)

. Can you guess what differentiates to give that (and then actually differentiate to see how well it works)?

Reply 3

JohnKennedy

yeah that was what i was thinking... so shall i let cosx = u

Reply 4

Glutamic Acid

But you'll still be left with a term of sin x. Is there another, better substitution?

let u= sinx?

Bear in mind that 2sin^2 (x) = something that makes matters much easier. (Think double angle formula)

Reply 7

Glutamic Acid

u = sin x will work. Daniel's method is what you'll probably use after some practice and the anti-derivatives are obvious.

aKarma

Bear in mind that 2sin^2 (x) = something that makes matters much easier. (Think double angle formula)

Huh? I can't see how considering the double angle formula will make this easier.

Reply 8

nota bene

aKarma

Bear in mind that 2sin^2 (x) = something that makes matters much easier. (Think double angle formula)

Huh? How does that simplify it?

Recognition is the by far easiest, and u=sinx if you're not able to.

Reply 9

aKarma

nota bene

Huh? How does that simplify it?

lol

Sorry about that, I hadn't really thought that through, just saw the 2sin^2 (x).

:dunce:

Reply 10

JohnKennedy

the result i get is 2sinx - (2/3)sin^3x + c

is that right folks?

Reply 11

nota bene

JohnKennedy

the result i get is 2sinx - (2/3)sin^3x + c

is that right folks?

No, where does the 2sinx come from? (plus there's still a sign error)

Reply 12

-G-a-v-

Think about the differential of sin^3 x....should look vaguely familiar :p:

The approach I'm thinking of is usually called 'integration by guesswork' or 'integration by inspection' - can be handy sometimes.

Reply 13

JohnKennedy

sorry i made a foolish mistake;

i now get (2/3)sin^3x + c

i think that's right...

Reply 14

Koudoo

Yep, you can solve it via exact integral, or you can rearrange 2sin²xcosx to -[(cos3x-cosx)/2]...

which integrates to...

-1/6 sin3x + 1/2 sinx + c

Which of course is exactly the same as (2/3)sin^3x + c

If Im not mistaken, that is a part of one of the integrating factor questions. On differential equations on FP1....