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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!
Use $$\frac{\sin^4x+\cos^4x}{\sqrt{1+\cos4x}}=\frac{1-\frac{1}{2}\sin^22x}{\sqrt2\cos2x}=\frac{\frac{1}{2}+\frac{1}{2}\cos^22x}{\sqrt2\cos2x}=\frac{\cos2x}{2\sqrt2(1-\sin^22x)}+\frac{1}{2\sqrt2}\cos2x=$$ $$=\frac{1}{4\sqrt2}\left(\frac{\cos2x}{1+\sin2x}+\frac{\cos2x}{1-\sin2x}\right)+\frac{1}{2\sqrt2}\cos2x.$$
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
I figured out why nobody is getting that answer: the question is wrong, the (correct) question is $$\int \frac{\cos^4 x-\sin^4 x}{\sqrt{1+\cos4x}} \ \text{d}x$$
$$\int { \frac { \cos ^{ 4 } x+\sin ^{ 4 } x }{ \sqrt { 1+\cos4x } } dx } =\frac { 1 }{ \sqrt { 2 } } \int { \frac { { \left( \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } \right) }^{ 2 }-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \sqrt { \cos ^{ 2 }{ 2x } } } dx } =\\ =\frac { 1 }{ \sqrt { 2 } } \int { \frac { 1-\frac { \sin ^{ 2 }{ 2x } }{ 2 } }{ \cos { 2x } } } dx=\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { 2-\sin ^{ 2 }{ 2x } }{ \cos { 2x } } } dx=\\ =\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { \cos ^{ 2 }{ 2x } +1 }{ \cos { 2x } } } dx=\frac { 1 }{ 2\sqrt { 2 } } \left[ \int { \cos { 2xdx } } +\int { \frac { dx }{ \cos { 2x } } } \right] =\\ =\frac { 1 }{ 4\sqrt { 2 } } \int { d\left( \sin { 2x } \right) } +\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { \cos { 2xdx } }{ \cos ^{ 2 }{ 2x } } } =\\ =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 4\sqrt { 2 } } \int { \frac { d\left( \sin { 2x } \right) }{ 1-\sin ^{ 2 }{ 2x } } } =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 8\sqrt { 2 } } \left[ \int { \frac { d\left( \sin { 2x } \right) }{ 1-\sin { 2x } } } +\int { \frac { d\left( \sin { 2x } \right) }{ 1+\sin { 2x } } } \right] =\\ =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 8\sqrt { 2 } } \left[ \int { \frac { d\left( 1+\sin { 2x } \right) }{ 1+\sin { 2x } } } -\int { \frac { d\left( 1-\sin { 2x } \right) }{ 1-\sin { 2x } } } \right] =\\ =\frac { \sin { 2x } }{ 4\sqrt { 2 } } +\frac { 1 }{ 8\sqrt { 2 } } \ln { \left| \frac { 1+\sin { 2x } }{ 1-\sin { 2x } } \right| + } C\\ $$
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.
