Question

I don't seem to be able to do this problem without looking at the answer lol
\[\int sin^{3}xcos^{2}x dx\]
do I try to eliminate either sin or cos?

Answer 1

Integral of sin^(3)(x)(1-sin^(2))

Answer 2

Now expand

Answer 3

integral of sin^(3)(x)dx - integral of sin^(5)(x)dx

Answer 4

\[\int sin^{3}(x)(1-sin^{2})\]

Answer 5

now you have two integrals

Answer 6

\[\int sin^{3}(x)dx - \int sin^{5}(x)dx\]

Answer 7

use substitution

Answer 8

Its better to go the other way

\[\int\limits_{?}^{?}\sin ^3 x \cos ^2 x = \int\limits_{?}^{?} sinx \cos^2 x ( 1 - \cos ^2x) dx\]

and then switch variables to u =cos x.

Answer 9

\[\int\limits_{}^{}\sin^{2}(x)\sin(x)dx - \int\limits_{}^{}\sin(x)(\sin^{2})^{2}dx\]

Answer 10

\[\int\limits_{}^{}(1-\cos^{2}(x))\sin(x)dx - \int\limits_{}^{}\sin(x)(1-\cos^{2}(x))^{2}\]

Answer 11

you can just keep splitting them up like this till the integral is solved

Answer 12

\[\int\limits_{}^{}\sin(x)dx - \int\limits_{}^{} \cos^{2}(x)\sin(x)dx + \int\limits_{}^{}\sin(x) - \int\limits_{}^{}\sin(x)(1-\cos^{2}(x))^{2}dx\]

Answer 13

your prof must really hate you lol

Answer 14

<falls off chair> there must be an easier way

Answer 15

soo tedious

Answer 16

I think I found something here I just can't get it in order

Answer 17

here is the instructions for dealing with integrals such as these provided in my calc 1 class

Answer 18

you have to do it this way

Answer 19

or at least this is the only way I know how to do it

Answer 20

here are*

Answer 21

u=cosx
du=-sinxdx
\[\int sin^{2}xcos^{2}xsinxdx \]

Answer 22

I would suggest doing the long simple method :) this isnt so bad eventually you will have the answer

Answer 23

You shouldnt be able to solve it like that lol.

sin3cos2 = sin 3 - sin 5
disregard the first term
sin5 = sin2sin3 = sin3 - sin3cos2
youre back at the start ahaha.

Answer 24

just do it my way trust me

Answer 25

\[\int (1-cos^{2})xcos^{2}xsinxdx\]

Answer 26

\[\int (1-cos^{2}x)cos^{2}xsinxdx\]

Answer 27

integrals can be long and tedious but you have to deal with it, at least they dont require complex methods

Answer 28

well at least this one doesnt

Answer 29

some do

Answer 30

u=cosx
du=-sinxdx
\[\int (1-cos^{2}x)cos^{2}xsinxdx\]
\[-\int (1-u^{2})u^{2}du\]

Answer 31

the time you spent messing with other techniques you could have finished this integral lol

Answer 32

Once again Its better to go the other way
\[ ∫sin^3xcos^2x=∫ sinxcos^2x(1−cos^2x)dx\]
and then switch variables to u =cos x.

du =-sin x dx so the integral becomes

\[\int\limits_{?}^{?} (-du) u^2(1-u^2) = \int\limits_{?}^{?} du (u^4 -u^2)\]

Answer 33

principle of nus tradamos

Answer 34

sigh ok

Answer 35

either way it's an integral, long and tidious

Answer 36

thanks everyone!

Answer 37

\[\frac{1}5 cos^{5}x -\frac 1 3 cos^{3} +C\]

Answer 38

its not so bad its very straight forward thus there isn't room for much mistake, also this document might help you through your adventures in the land of calc 1

Answer 39

it covers pretty much everything covered in calc 1 at my university very briefly

Answer 40

it is where that image came from

Answer 41

i see, thanks @Australopithecus

Answer 42

Is this in your standrad table of integrals ?

\[\int\sin^m(x)\cos^n(x)\text dx\]\[=\frac{\sin^{m+1}(x)\cos^{n-1}}{m+n}+\frac{n-1}{m+n}\int\sin(x)^m\cos^{n-2}(x)\text dx\]

Answer 43

lets see
\[\int\sin^3(x)\cos^2(x)\text dx\]
\[=\frac{\sin^{4}(x)\cos(x)}{5}+\frac{1}{5}\int\sin(x)^3\text dx\]

Answer 44

something seems off though

Answer 45

the answer should be in cosines

Answer 46

\[\frac 1 5 cos^ 5-\frac 1 3 cos^ 3 x + C\]

Answer 47

some trig substitution

Answer 48

that would make the last term still \[\frac 1 {20} sin^4\]

Answer 49

I would think?

Answer 50

hmm