Question

How to integrate sin2xcos2x/sqrt(1+cos4x)?

I know that if I do anti-differentiation, I would get 1+cos4x to have the differential sin2xcos2x, but I can't seem to figure out how 1+cos4x differentiate to get sin2xcos2x

Answer 1

\[\int\limits_{}^{}\frac{ \sin(2x)\cos(2x) }{ \sqrt{(1+\cos(4x)} } dx\]

Answer 2

is this ur question ?

Answer 3

\(\color{#000000 }{ \displaystyle \int\frac{\sin(2x)\cos(2x)}{\sqrt{1+\cos(4x)}}dx }\)

You know that,
\(\color{#000000 }{ \displaystyle\sin 2z = 2\sin z \cos z }\)
(In your case, \(\color{#000000 }{ \displaystyle z:=2x }\))

\(\color{#000000 }{ \displaystyle \int\frac{2\sin(4x)}{\sqrt{1+\cos(4x)}}dx }\)

Answer 4

Then, substitute; \(\color{#000000 }{ \displaystyle p=1+\cos(4x) }\)

(Or, you can use any other latter/symbol that you want)

Answer 5

But sin 2x=2sin 2xcos2x
\[2\sin2xcos2x \times \cos2x=(2\sin2x)(2\cos2x)?\]

Sorry, but I'm still a bit confused how it would become 2 sin(4x) @SolomonZelman

Answer 6

\(\color{#000000 }{ \displaystyle \sin(2z)=2\sin z \cos z }\)

That is the rule, right?

Answer 7

Yes, I made a mistake there

Answer 8

\(\color{#000000 }{ \displaystyle \int\frac{\sin(2x)\cos(2x)}{\sqrt{1+\cos(4x)}}dx }\)

You know that,
\(\color{#000000 }{ \displaystyle\sin 2z = 2\sin z \cos z }\)
(In your case, \(\color{#000000 }{ \displaystyle z:=2x }\))

\(\color{#000000 }{ \displaystyle \int\frac{1}{2}\frac{\sin(4x)}{\sqrt{1+\cos(4x)}}dx }\)

Answer 9

The rule is:
\(\color{#000000 }{ \displaystyle \sin(2z)=2\sin z \cos z }\)

When you substitute \(\color{#000000 }{ \displaystyle 2x }\) instead of \(\color{#000000 }{ \displaystyle z }\) into the equation,
you are going to get the following:

\(\color{#000000 }{ \displaystyle \sin(2\color{#ff0000}{\cdot 2x})=2\sin \color{#ff0000}{ (2x)} \cos \color{#ff0000}{ (2x)} }\)
\(\color{#000000 }{ \displaystyle \sin(4x)=2\sin (2x) \cos (2x) }\)

And therefore,
\(\color{#000000 }{ \displaystyle (1/2)\cdot \sin(4x)=\sin (2x) \cos (2x) }\)

Answer 10

So, your integral is;

\(\color{#000000 }{ \displaystyle \frac{1}{2}\int\frac{\sin(4x)}{\sqrt{1+\cos(4x)}}dx }\)

And that can be solved by setting
\(\color{#000000 }{ \displaystyle p=1+\cos(4x) }\)

Answer 11

Do you have questions about this?

Answer 12

@SolomonZelman u are very good with integration and even with explanation :)

Answer 13

thanks ]]

Answer 14

Thank you!! @SolomonZelman. I get it now. :)

Answer 15

What did you get for the integral?

Answer 16

(what is your answer? if you have an answer now...)

Answer 17

Although I was just about to ask you. The answer is \[-\frac{ 1 }{ 4 }\sqrt{1+\cos (4x)}+c\]

Answer 18

And carrying on from where you left off, I arrived at the integral of 1/8 sqrt (1+cos (4x)), because if we consider the 1/2 left off from the previous steps...

Answer 19

\(\color{#000000 }{ \displaystyle \int\frac{\sin(2x)\cos(2x)}{\sqrt{1+\cos(4x)}}dx }\)

You know that, \(\color{#000000 }{ \displaystyle\sin 2z = 2\sin z \cos z }\) (in your case, \(\color{#000000 }{ \displaystyle z:=2x }\))

\(\color{#000000 }{ \displaystyle \frac{1}{2}\int\frac{\sin(4x)}{\sqrt{1+\cos(4x)}}dx }\)

\(\color{#000000 }{ \displaystyle p=1+\cos(4x) }\)
\(\color{#000000 }{ \displaystyle dp=-4\sin(4x)dx~~\Longrightarrow~~(-1/4)dp=\sin(4x) dx }\)

\(\color{#000000 }{ \displaystyle -\frac{1}{8}\int\frac{1}{\sqrt{~p~}}dp }\)

\(\color{#000000 }{ \displaystyle -\frac{1}{8}\int p^{-1/2}dp }\)

\(\color{#000000 }{ \displaystyle -\frac{1}{4}p^{1/2} + {\rm C} }\)

\(\color{#000000 }{ \displaystyle -\frac{1}{4}\sqrt{1+\cos(4x)}+ {\rm C} }\)

Answer 20

yes, I got the same answer

Answer 21

Oh, yes. sorry, my careless mistake. I didn't divide by 1/2 (ie multiply by 2) when I was doing the integral of 1+cos(4x).
Thank you very much.

Answer 22

I don't really see any mistakes that you made ... all I saw is that you arrived at the same answer as I did....