Question

integrate sin 2x / √(1 + cos² x) dx

Answer 1

Hmm, try substitute
\[u=\cos^2(x) \\ \\du=-2\cos(x)\sin(x)\]
\[-\int\limits \frac{1}{\sqrt{u+1}} \, du\]\
Substitute one more time,
\[t=u+1 \\ \\dt=du\]
\[-\int\limits \frac{1}{\sqrt{t}} \, dt\]
Can you do it now?

Answer 2

Note: from starting, I used sin(2x)=2sin(x)cos(x)

Answer 3

i dont really get it :(

Answer 4

Thought so, :D I'll rearrange it

Answer 5

in my book it says that the answer is -2 √(1 + cos² x) + c

Answer 6

You have
\[\int\limits \frac{\sin 2x }{\sqrt{ \cos ^2x+1}} \, dx\]
Change the sin(2x) to 2sin(x)cos(x), That's an identity.
\[\int\limits \frac{2sin(x)cos(x) }{\sqrt{ \cos ^2x+1}} \, dx\]
Then, let
\[u=\cos^2(x) \\ \\du=-2\cos(x)\sin(x) dx \\ \\-du=2\sin(x)\cos(x)dx\]
\[\int\limits \frac{-du }{\sqrt{ u+1}} \, \]
Substitute one more time, Let
\[t=u+1 \\ \\dt=du\]
\[\int\limits \frac{-dt }{\sqrt{ t}} \, \]
\[-\int\limits \frac{1 }{\sqrt{ t}} dt\, \]
\[-\int\limits t^{-\frac{1}{2}}dt\, \]
Integrate it,
\[-2 \sqrt{t}+c\]
Substitute back,
\[-2 \sqrt{u+1}+c\]
\[-2 \sqrt{cos^2x+1}+c\]

Answer 7

i see...
but in my level, we haven't learn how to differentiate cos² x directly, we only learn how to differentiate cos x
so i dont really get it
is there any other way?

Answer 8

I don't think so, but differentiating cos² x is using chain rule, you should know that before doing integration. Hmm, that's weird.

Answer 9

okay i'll try. thank you!

Answer 10

i only didnt undertsand that part but the rest i do so thanks a lot !