Question

integral of cosx/sinx dx

Answer 1

Use u-substitution

Answer 2

Use that derivative formula instead.

Answer 3

cosx*sinx^-1

if u is sinx then derivative of u is cosx

cosx is there so sinx^-1 plus 1 equals 0...so is it just zero??? i think thats wrong

Answer 4

Isn't cosx/sinx just cotx?

Answer 5

omg yeah

Answer 6

You're right~!

Answer 7

Use subtitiution and you will get ln|sinx|+C

Answer 8

Which is what you get when you integrate cotx

Answer 9

so integral of cotx is lnsinx

Answer 10

thanks so much

Answer 11

I'm surprised I was the one to spot that first... I'm normally the slowest one xD

Answer 12

:)

Answer 13

thank you

Answer 14

So u-sub WOULD work.

Answer 15

would anyone know how to do integral of cos^32xsin2x dx

Answer 16

Oh that is also a good point to bring up saifoo. That is where the integral of cotx was derived so yeah ^^

Answer 17

thats integral of cos cubed 2x (sin2x)

Answer 18

\[\int\limits \cos^3x(2xsin(2x))dx\]?

Answer 19

Ohh nevermind my parenthesis is off.

Answer 20

yeah but no x after the cos cubd...theres a 2x after the cos cubed...yeah haha parenthesis off

Answer 21

Yup. Thanks. @smokeydabear

@Jhannybean 's right too about substitution, but that is a pretty long method i believe.

Answer 22

\[\int\limits \cos^32x(\sin(2x))dx\] here you can solve integrating by parts

Answer 23

Substitution would have given you a very similar answer, and it wouldn't be long if you noticed that the top is the derivative of the bottom @saifoo.khan

Answer 24

Use ILATE for u-sub. In reference that means Inverse functions> logarithmic > arithmetic > trigonometric> exponential . It's a simple method when dealing with integrating by parts

Answer 25

ok thansk

Answer 26

@Jhannybean : How can we apply ILATE principle here? both terms are trigonometric terms.

Answer 27

I wasn't referencing this problem haha, I was stating when using integration by parts, ILATE is a helpful formula. :(

Answer 28

Ohh, rightt. That was confusing me.
By parts is great too.