Question

Evaluating integrals. Trig! :D

Answer 1

First replace sin2x with 2sinxcosx.

Let u = cosx.

Answer 2

Is that an identity?

Answer 3

Yeah... it's a pretty commonly used one.

Answer 4

oh. Pardon my forgetfulness xD

Answer 5

u = cos(x)
du=sin(X)

Answer 6

well, du = sin(x) dx

Answer 7

Derivative of cosx is -sinx

Answer 8

right right.

Answer 9

....

Answer 10

If we make
u = 21+cos^2(x)
du = sin2(x) + (x/2)
isn't that easier to work with?

Answer 11

Sure that works too, u=21+cos^2x.

But your du is very wrong in that case.

Answer 12

\[\frac{ \frac{ \sin2x }{ 2 } +x}{ 2 }\] no? o.o

Answer 13

That's even further from correct. Use the chain rule.

The derivative of \[\large f[g(x)]\] is \[\large f'[g(x)]*g'(x)\]

Answer 14

...what' the g(x) in this case? just (x)?

Answer 15

g is cosx, since your function is (cosx)^2.

Answer 16

du = 2cos(x)-sin(x)

Answer 17

Be careful how you write that. It looks like subtraction.

Answer 18

oops heh
-sin(x)*2cos(x)

Answer 19

Better. Now it should be easier to finish, easier than it would've been with my original suggestion of u=cosx.

Answer 20

-du=2sin(x)cos(x)
\[-\int\limits_{}^{}\frac{ 1 }{ u }*du\]

Answer 21

= -ln|u|+C
= -ln|21+cos^2(x)|+C
= -ln(21+cos^2(x))+C

Answer 22

Good job.

Answer 23

Yay!! thank you so much :)

Answer 24

You're welcome :)