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.Sam. Group TitleBest ResponseYou've already chosen the best response.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?
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.1
Note: from starting, I used sin(2x)=2sin(x)cos(x)
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
i dont really get it :(
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.1
Thought so, :D I'll rearrange it
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
in my book it says that the answer is 2 √(1 + cos² x) + c
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.1
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, \[2 \sqrt{u+1}+c\] \[2 \sqrt{cos^2x+1}+c\]
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.1
I don't think so, but differentiating cos² x is using chain rule, you should know that before doing integration. Hmm, that's weird.
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
okay i'll try. thank you!
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
i only didnt undertsand that part but the rest i do so thanks a lot !
 one year ago
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