Question

integrate cos^3t sin^4t

Answer 1

use u-substitution

Answer 2

@RolyPoly

Answer 3

\[\int cos^3t sin^4tdt\]\[=\int cost(cos^2t) sin^4tdt\]\[=\int(cos^2t) sin^4td(sint)\]\[=\int(1-sin^2t) sin^4td(sint)\]\[=\int (sin^4t-sin^6t)d(sint)\]\[=...\]

Answer 4

den it become..sin^5t - sin^7t

Answer 5

how do we solve that ? :P...sorry i am totaly weak in this :( @RolyPoly

Answer 6

My bad!
Use u-sub then...
Let u=sint
What is du/dt?

Answer 7

cost

Answer 8

Yup.
So, du = cost dt
cos^3t = cost (cos^2 t)
Use identity sin^2 t + cos^2 t = 1
cos^2 t can be rewritten as 1-sin^2 t
Understand so far?

Answer 9

i guess you made a mistake at the previous derivation...please check !

Answer 10

May I know where I made mistakes?

Answer 11

∫cost(cos^2t)sin4tdt

∫(cos^2t)sin4t(sint)

where did the cost t go ?

Answer 12

The dt changed to d(sint) .
Since d(sint) = cost dt

Answer 13

oh!! ...

Answer 14

Got it?

Answer 15

it then becomes integrate of U^5t-U^7t

Answer 16

Not really.
∫(cos^2t)sin^4t d(sint)
=∫(1-sin^2t)sin^4t d(sint)
=∫sin^4t - sin^6t d(sint)
Let u = sint
=∫u^4 - u^6 du
=... ?

Answer 17

and then..u^5/5 - u^7/7

Answer 18

Yes.
And put back u=sint into that one.
u^5/5 - u^7/7
= ... ?

Answer 19

awesomenss...:D :D ..and i have another one..integration is so hard :(

Answer 20

Couldn't agree more!