Question

Help with integration problem. Question is attached!

Answer 1

int by subs
let u=49+cos^2 x

Answer 2

So i set up u=cosx and du = -sinx dx. And then -du=sinx dx.

Answer 3

I did a second method but I think it might be longer. But you could use trig identity and then u-sub twice.

Answer 4

it can be int(-du/u)

Answer 5

\[∫\frac{ 2\sin(x)\cos(x) }{ 49+\cos^2(x) }dx\]

Answer 6

So then my integral became -2 times the integral of (sinx/ 49+cosx^2) dx

Answer 7

@abb0t it says math processing error so i cant see what u posted

Answer 8

\[u = 49 + \cos^2(x) \]

Answer 9

\[\int\limits \frac{ 2\sin(x)\cos(x) }{ 49+\cos^2(x) }dx\]

better?

Answer 10

Then it turned to -2 times the integral of (du/49+u^2). Im not sure what to do now.

Answer 11

Let me refresh the page to see if i can see ur equatons

Answer 12

YES I can see them now!

Answer 13

oh i used a different u. I had used u=cosx. But ok lets see what ur answer gives me or if its easier to do.

Answer 14

use trig identity: sin(2x) = 2sin(x)cos(x)

Answer 15

Ok and what happens to the denominator with the 49+cosx^2

Answer 16

that's your "u"
take the derivative and see what you get.
this is a u-substitution problem.

Answer 17

so du=-cosx^2 OR -du = cosx^2. And so does the integral become -integral of (u/ 49+du)

Answer 18

\[\int\limits \frac{ 2\sin(x)\cos(x) }{ 49+\cos^2(x) }dx\]
\[u = 49 + \cos^2(x)\]
\[du = -2\cos(x)\sin(x)\]
\[-du = 2\cos(x)\sin(x)\]
\[-\int\limits \frac{ 1 }{ u }dx\]

Answer 19

WAIT NO! ITS ln(x) SORRY

Answer 20

OR -ln(u) right??

Answer 21

is -ln(49+cosx^2) the answer??

Answer 22

yes, the finally be
-ln(u) + c
substitute back that u=49+cos^2 x

Answer 23

the last one, is correct

Answer 24

Yep.

Answer 25

Thanks for confirming my answer.

Answer 26

fa sho homie.

Answer 27

Sorry just one question. Since sin(2x) = 2sin(x)cos(x) and -du = 2cos(x)sin(x) is that the same thing? Also how did the last integral turn to 1/udu??

Answer 28

im confused where the one came

Answer 29

It is the same thing. I just rearraged it.
I/u is because u is my substitution. and my du = 2sin(x)cos(x)

Answer 30

\[\frac{ du }{ u } = \frac{ 2\sin(x)\cos(x) }{ 49+\cos^2(x) }\]

Answer 31

Oh ok. Thanks. Yeah i was just confused if sin(2x) = 2sin(x)cos(x) and 2cos(x)sin(x) were equivalent.

Answer 32

i see it now. Thank u for clarifying.

Answer 33

Same thing as 2xy = 2yx

Answer 34

hope it does make sense