Question

integrate
sin^4xcos^5xdx

Answer 1

Note that:

∫ sin^3(x) * cos^4(x) dx
= ∫ sin^2(x) * cos^4(x) [sin(x) dx]
= ∫ [1 - cos^2(x)] * cos^4(x) [sin(x) dx]
= - ∫ [1 - cos^2(x)] * cos^4(x) [-sin(x) dx].

Then, letting u = cos(x) <==> du = -sin(x) dx yields:

- ∫ [1 - cos^2(x)] * cos^4(x) [-sin(x) dx]
= - ∫ (1 - u^2) * u^4 du
= - ∫ u^4 - u^6 du
= ∫ u^6 - u^4 du
= ∫ u^6 du - ∫ u^4 du
= (1/7)u^7 - (1/5)u^5 + C
= (1/7)cos^7(x) - (1/5)cos^5(x) + C. <== ANSWER

I hope this helps!

Answer 2

hihi

Answer 3

he looks right

Answer 4

thank you so much

Answer 5

@Rohangrr what happen to the other sin x and cosx

Answer 6

still there? have any questions?

Answer 7

yes im still here
I was just confused where was the other sinx cosx

Answer 8

looking at rohangff work, what line?

Answer 9

in that first line where sin^(3)x cos^(4)x dx

did he factor that out then what happen to the other one.?

Answer 10

hmm, you're right... i don't think he did your problem... double checking...

Answer 11

I really dont know how to integrate this what formula did he used?

Answer 12

yea, he didn't do the problem but the method should be the same...

Answer 13

ok, lemme try..

Answer 14

thank you so much then

Answer 15

you follow up to that last line?