Are These Trigonometric Integral Solutions Equivalent?

[paulmdrdo1]

i was thinking hard how to integrate this, but none of the techniques I know did work.
please kindly help with this matter. thanks!

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

 

[alyafey22]

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

 

[paulmdrdo1]

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

 

[paulmdrdo]

you can use this identity $\cos(2a)=2\cos^2(a)-1$ it's called half-angle indentity.

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

good luck sa'yo!

 

[soroban]

Hello, paulmdrdo!

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

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

. . Hence: .\sqrt{1+\cos\theta} \:=\:\sqrt{2}\cos\tfrac{\theta}{2}The integral becomes: .\displaystyle \sqrt{2}\int \cos\tfrac{\theta}{2}\,d\theta

Got it?

 

[paulmdrdo1]

Hello, paulmdrdo!


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

. . Hence: .\sqrt{1+\cos\theta} \:=\:\sqrt{2}\cos\tfrac{\theta}{2}The integral becomes: .\displaystyle \sqrt{2}\int \cos\tfrac{\theta}{2}\,d\theta

Got it?

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?

 

[Prove It]

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?

Yes they are equivalent.

\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*}