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# intergration watch

1. how do i intergrate

sin2xcosx ?

thanks
2. There's a thread just below this one of exactly the same nature.
3. sin2x = 2sinxcosx

=> (2) INT (sinx)(cosx)^2 dx

= (-2) INT (-sinx)(cosx)^2 dx

Now use the fact that: INT [f '(x)].[f(x)]n dx = [1/(n+1)][f(x)]n+1 + c

= (-2/3)cos3 + c
4. (Original post by sarah12345)
how do i intergrate

sin2xcosx ?

thanks
is this perceived as being the hardest integral or what - quite a few threads asking about that particular question

use fact that:INT [f '(x)].[f(x)]n dx = [1/(n+1)][f(x)]n+1 + c, as said by mockel - its basically recognition - a skill you'll pick up with practice

pk
5. i got one quite complicated
intergate cosx*sqrt(cos2x)
6. (Original post by n_e_m)
i got one quite complicated
intergate cosx*sqrt(cos2x)
Write cos2x = 1 - 2sin^2 x and then sub u = sinx

Then write u = 1/rt(2) sint say
7. (Original post by n_e_m)
i got one quite complicated
intergate cosx*sqrt(cos2x)
cos2x = 1 - 2sin^2x

Let u = sinx -> 2u^2 = 2sin^2x -> 1 - 2u^2 = 1 - 2sin^2x
Also: du/dx = cosx -> dx = 1/cosx du

-> Int. (cosx).Sqrt[cos2x] dx = Int. cosx.Sqrt[1 - 2u^2].(1/cosx) du. = Int. Sqrt(1 - u^2) du. = Int. (1 - u^2)^(1/2) du. = [-1/(2u)](2/3)(1 - u^2)^(3/2) + k = -[1/(3u)](1 - u^2)^(3/2) + k = -[1/(3sinx)][cos^2x]^(3/2) + k = -[1/(3sinx)][cos^3x] + k = (-1/3){[cos^2x.cosx]/sinx} + k = (-1/3)(cos^2x/tanx) + k
8. many thanx...
i wonder if this type of complicated integration could come up on P5 paper?

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