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How do we integrate the following?

$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$

I tried to simplify this, but I cannot seem to proceed further than the below form:

$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos2x}dx}}$

$\implies \frac{1}{2\sqrt2}\log |\sec 2x + \tan 2x| + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos^2x-\sin^2x}dx} + C$

The answer that I'm supposed to get is:

$\frac{x}{\sqrt2}+C$

Please help, thanks!

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5 Answers 5

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Use $$\frac{\sin^4x+\cos^4x}{\sqrt{1+\cos4x}}=\frac{1-\frac{1}{2}\sin^22x}{\sqrt2\cos2x}=\frac{\frac{1}{2}+\frac{1}{2}\cos^22x}{\sqrt2\cos2x}=\frac{\cos2x}{2\sqrt2(1-\sin^22x)}+\frac{1}{2\sqrt2}\cos2x=$$ $$=\frac{1}{4\sqrt2}\left(\frac{\cos2x}{1+\sin2x}+\frac{\cos2x}{1-\sin2x}\right)+\frac{1}{2\sqrt2}\cos2x.$$

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  • $\begingroup$ I'm sorry, I made an error in the question... Let me rectify it. $\endgroup$ Feb 3, 2018 at 18:32
  • $\begingroup$ There! The $1 + \cos4x$ in the denominator was supposed to be under a root $\endgroup$ Feb 3, 2018 at 18:37
  • $\begingroup$ I added something. See now. $\endgroup$ Feb 3, 2018 at 18:39
  • $\begingroup$ Shouldn't the simplification of $\frac{\frac{1}{2}+\frac{1}{2}\cos^22x}{\sqrt2\cos2x}$ be $\frac{1 + \cos^22x}{2\sqrt2\cos2x}$? Which would then give $\frac{1}{2\sqrt{2}\cos2x} + \frac{cos2x}{2\sqrt{2}}$? $\endgroup$ Feb 3, 2018 at 18:51
  • $\begingroup$ Yes, but see also the rest. $\endgroup$ Feb 3, 2018 at 18:52
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Using $\cos2x=1-2\sin^2x=2\cos^2x-1,$

$$I=\dfrac{\sin^4x+\cos^4x}{\sqrt{1+\cos4x}}=\dfrac{(1-\cos2x)^2+(1+\cos2x)^2}{4\sqrt2|\cos2x|}=\dfrac{1+\cos^22x}{2\sqrt2|\cos2x|}$$

For $\cos2x>0,$

$$2\sqrt2I=\sec2x+\cos2x$$

Now use Integral of the secant function

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I figured out why nobody is getting that answer: the question is wrong, the (correct) question is $$\int \frac{\cos^4 x-\sin^4 x}{\sqrt{1+\cos4x}} \ \text{d}x$$

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$$\int { \frac { \cos ^{ 4 } x+\sin ^{ 4 } x }{ \sqrt { 1+\cos4x } } dx } =\frac { 1 }{ \sqrt { 2 } } \int { \frac { { \left( \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } \right) }^{ 2 }-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \sqrt { \cos ^{ 2 }{ 2x } } } dx } =\\ =\frac { 1 }{ \sqrt { 2 } } \int { \frac { 1-\frac { \sin ^{ 2 }{ 2x } }{ 2 } }{ \cos { 2x } } } dx=\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { 2-\sin ^{ 2 }{ 2x } }{ \cos { 2x } } } dx=\\ =\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { \cos ^{ 2 }{ 2x } +1 }{ \cos { 2x } } } dx=\frac { 1 }{ 2\sqrt { 2 } } \left[ \int { \cos { 2xdx } } +\int { \frac { dx }{ \cos { 2x } } } \right] =\\ =\frac { 1 }{ 4\sqrt { 2 } } \int { d\left( \sin { 2x } \right) } +\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { \cos { 2xdx } }{ \cos ^{ 2 }{ 2x } } } =\\ =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 4\sqrt { 2 } } \int { \frac { d\left( \sin { 2x } \right) }{ 1-\sin ^{ 2 }{ 2x } } } =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 8\sqrt { 2 } } \left[ \int { \frac { d\left( \sin { 2x } \right) }{ 1-\sin { 2x } } } +\int { \frac { d\left( \sin { 2x } \right) }{ 1+\sin { 2x } } } \right] =\\ =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 8\sqrt { 2 } } \left[ \int { \frac { d\left( 1+\sin { 2x } \right) }{ 1+\sin { 2x } } } -\int { \frac { d\left( 1-\sin { 2x } \right) }{ 1-\sin { 2x } } } \right] =\\ =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 8\sqrt { 2 } } \ln { \left| \frac { 1+\sin { 2x } }{ 1-\sin { 2x } } \right| + } C\\ $$

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  • $\begingroup$ Nicely done! (+1) $\endgroup$ Feb 3, 2018 at 18:52
  • $\begingroup$ This is a pretty good, but how am I supposed to simplify this to the given answer? $\endgroup$ Feb 3, 2018 at 19:00
  • $\begingroup$ you can simplify $\sin { 2x } =2\sin { x } \cos { x } $ or,you go further as $\ln { \left| \frac { 1+\sin { 2x } }{ 1-\sin { 2x } } \right| = } \ln { \left| \frac { \sin ^{ 2 }{ x } +2\sin { x\cos { x } +\cos ^{ 2 }{ x } } }{ \sin ^{ 2 }{ x } -2\sin { x\cos { x } +\cos ^{ 2 }{ x } } } \right| = } \\ =\ln { \left| \frac { { \left( \sin { x } +\cos { x } \right) }^{ 2 } }{ { \left( \sin { x } -\cos { x } \right) }^{ 2 } } \right| = } \ln { \left| \frac { \sin { x+\cos { x } } }{ \sin { x } -\cos { x } } \right| } $ $\endgroup$
    – haqnatural
    Feb 3, 2018 at 19:05
  • $\begingroup$ I did that one too, but that's no where close the required answer... but I guess it is something as I couldn't even integrate in earlier :D $\endgroup$ Feb 3, 2018 at 19:18
  • $\begingroup$ what should the required answer be? $\endgroup$
    – haqnatural
    Feb 3, 2018 at 19:22
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If $\cos 2x > 0$ and the integrand is wrong as Rudr Pratap Singh points out, then: \begin{align} \frac{\cos^4 x-\sin^4 x}{\sqrt{1+\cos 4x}} &= \frac{(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)}{\sqrt{2\cos^2 2x}} \\ &= \frac{(1)(\cos 2x)}{\sqrt2|\cos2x|} \\ &= \frac 1 {\sqrt2}. \end{align} Hence, $$\int \frac{\cos^4x-\sin^4x}{\sqrt{1+\cos 4x}} \ \text dx = \frac x {\sqrt2} +c,$$ as desired by O.P.

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