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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

please kindly help with this matter. thanks!

$\displaystyle\int\sqrt{1+\cos\theta}d\theta$

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

please kindly help with this matter. thanks!

$\displaystyle\int\sqrt{1+\cos\theta}d\theta$

- Jan 17, 2013

- 1,667

You should get rid of the root . A useful trigonometric identity might work here .

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- #3

- May 13, 2013

- 386

what trig identity? i only know few basic identities. none of them did work.

- Aug 7, 2013

- 21

or simply, you can multiply and divide the integrand to $\sqrt{1-\cos\theta}$

good luck sa'yo!

$\displaystyle\int\sqrt{1+\cos\theta}\,d\theta$

Identity: .[tex]\cos^2\!\tfrac{\theta}{2} \:=\:\frac{1+\cos\theta}{2} \quad \Rightarrow \quad 1 + \cos\theta \:=\:2\cos^2\!\tfrac{\theta}{2}[/tex]

. . Hence: .[tex]\sqrt{1+\cos\theta} \:=\:\sqrt{2}\cos\tfrac{\theta}{2}[/tex]

The integral becomes: .[tex]\displaystyle \sqrt{2}\int \cos\tfrac{\theta}{2}\,d\theta[/tex]

Got it?

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- #6

- May 13, 2013

- 386

soroban, i solved it on paper but i don't have time to post the solution,Hello, paulmdrdo!

Identity: .[tex]\cos^2\!\tfrac{\theta}{2} \:=\:\frac{1+\cos\theta}{2} \quad \Rightarrow \quad 1 + \cos\theta \:=\:2\cos^2\!\tfrac{\theta}{2}[/tex]

. . Hence: .[tex]\sqrt{1+\cos\theta} \:=\:\sqrt{2}\cos\tfrac{\theta}{2}[/tex]

The integral becomes: .[tex]\displaystyle \sqrt{2}\int \cos\tfrac{\theta}{2}\,d\theta[/tex]

Got it?

here's my answer.. $\displaystyle 2\sqrt{2}\sin\left(\frac{\theta}{2}\right)+C$ is this correct?

and also, i tried to solve by rationalizing the integrand the result is $2\sqrt{1-\cos\theta}+C$

By any chance are my answers equivalent? how do I know that the different forms of my answers were the same?

Yes they are equivalent.soroban, i solved it on paper but i don't have time to post the solution,

here's my answer.. $\displaystyle 2\sqrt{2}\sin\left(\frac{\theta}{2}\right)+C$ is this correct?

and also, i tried to solve by rationalizing the integrand the result is $2\sqrt{1-\cos\theta}+C$

By any chance are my answers equivalent? how do I know that the different forms of my answers were the same?

[tex]\displaystyle \begin{align*} \cos{ \left( 2\theta \right) } &\equiv 1 - 2\sin^2{(\theta)} \\ \cos{ \left( x \right) } &\equiv 1 - 2\sin^2{ \left( \frac{x}{2} \right) } \textrm{ if } x = 2\theta \\ 2\sin^2{ \left( \frac{x}{2} \right) } &\equiv 1 - \cos{ \left( x \right) } \\ \sin^2{ \left( \frac{x}{2} \right) } &\equiv \frac{1 - \cos{ \left( x \right) } }{2} \\ \sin{ \left( \frac{x}{2} \right) } &\equiv \frac{ \sqrt{ 1 - \cos{(x)} } }{ \sqrt{2} } \\ 2\sqrt{2}\sin{ \left( \frac{x}{2} \right) } &\equiv 2\sqrt{ 1 - \cos{(x)}} \end{align*}[/tex]