## shaqadry one year ago integrate sin 2x / √(1 + cos² x) dx

1. .Sam.

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?

2. .Sam.

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

i dont really get it :(

4. .Sam.

Thought so, :D I'll rearrange it

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

6. .Sam.

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, $-2 \sqrt{u+1}+c$ $-2 \sqrt{cos^2x+1}+c$

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?

8. .Sam.

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