Question

half-angle formula:
I'm trying to integrate the sqrt 2(1+cosx) and I know you can use half-angle to turn it into sqrt 4(cos^2(x)). But I don't see *how* it turned into that :( Can someone please explain

Answer 1

I think it's the second identity in this link:
http://www.freemathhelp.com/images/halfangles.png
and I think you made a typo

Answer 2

\[\color{red}{\int \sqrt{2(1+\cos(x))}dx}\]\[=\int \sqrt{2} \cdot \sqrt{1+\cos(x)}dx\]\[=\sqrt{2}\int\sqrt{1+\cos(x)}dx\]If \(\cos(2x) =2\cos^2(x)-1\), then \(\cos(x) = 2\cos^2\left(\dfrac{x}{2}\right)-1\implies \color{blue}{\cos(x)+1=2\cos^2\left(\dfrac{x}{2}\right)} \)\[=\sqrt{2}\int \sqrt{2\cos^2\left(\dfrac{x}{2}\right)}dx\]\[=\sqrt{2}\int \sqrt{2}\sqrt{\cos^2\left(\dfrac{x}{2}\right)}dx \]\[=2\int\sqrt{\cos^2\left(\dfrac{x}{2}\right)}dx\]Let \(u=\dfrac{x}{2} ~,~ du = \dfrac{1}{2}dx \implies 2du = dx\)

Answer 3

\[=4\int \sqrt{\cos^2(u)}du\]\[4\int \cos(u)du =\color{red}{4\sin(u)+C} = 4\sin\left(\dfrac{x}{2}\right)+C\]Can be simplified one more step but I'll let you figure that out.

Answer 4

\[\int\sqrt {4(cos^2(x/2))} \]

Answer 5

dx

Answer 6

It's good

Answer 7

For the second one \[2 \int\limits \sqrt{\cos^2(x/2)}dx = 4 \int\limits \sqrt{\cos^2u}du\] letting u = x/2 and du = 1/2 dx you can finish it off :)

Answer 8

About 1-2 steps away hehe

Answer 9

Water u talkin' bout mate?

Answer 10

The second integral

Answer 11

Oh wait, is it suppose to be the same one haha?

Answer 12

Yeah the squared and the square root cancel eachother out.

Answer 13

I was doing what dan posted

Answer 14

Yeah it's the same one, hence the equal sign lmao -.-

Answer 15

Yeah ok that makes sense lol I just realized that, because I just went over your work

Answer 16

then dan wrote that

Answer 17

My post was just lagging really bad and I was afraid it was going to disappear like the first time D::

Answer 18

:)