Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$

Question

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!

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

$$ \begin{aligned}\int \frac{\cos ^4 x+\sin ^4 x}{\sqrt{1+\cos 4 x}} d x = & \int \frac{\left(\cos ^2 x+\sin ^2 x\right)^2-2 \cos^2 x \sin ^2 x}{\sqrt{2 \cos ^2 2 x}} d x \\ = & \frac{1}{\sqrt{2}} \int \frac{1-\frac{1}{2} \sin ^2(2 x)}{\cos (2 x)} d x \\ = & \frac{1}{\sqrt{2}} \int \frac{1-\frac{1}{2}\left(1-\cos ^2(2 x)\right)}{\cos (2 x)} d x \\ = & \frac{1}{2 \sqrt{2}}\left[\int \sec (2 x) d x+\int \cos (2 x) d x\right] \\ = & \frac{1}{4 \sqrt{2}}[\ln |\sec (2 x)+\tan (2 x)|+\sin (2 x)]+C . \end{aligned} $$

Answer 5

With the substitution

$$y=\sec x+\tan x \implies x = 2\arctan\frac{y-1}{y+1} = 2\left(\arctan y-\frac\pi4\right)\implies dx = \frac{2\,dy}{y^2+1}$$

and partial fraction decomposition, we can break it up into simpler pieces:

$$\begin{align*} & \int \frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos(4x)}} \, dx \\ &= \int \frac{\frac{16y^4+\left(y^2-1\right)^4}{\left(y^2+1\right)^4}}{\sqrt{1 + \frac{y^8 - 28y^6 + 70y^4 - 28y^2 + 1}{\left(y^2+1\right)^4}}} \, \frac{2\,dy}{y^2+1} \\ &= \sqrt2 \int \frac{y^8-4y^6+22y^4-4y^2+1}{y^4-6y^2+1} \, \frac{dy}{(y^2+1)^3} \\ &= \frac1{2\sqrt2} \int \left(\frac{16}{\left(y^2+1\right)^3} - \frac{16}{\left(y^2+1\right)^2} - \frac2{y^2+1} + \frac{y-1}{y^2-2y-1} - \frac{y+1}{y^2+2y-1}\right) \, dy \\ &= \frac1{16\sqrt2} \left(\frac{4y\left(1-y^2\right)}{\left(1+y^2\right)^2} + \ln \left|y^2-2y-1\right| - \ln \left|y^2+2y-1\right|\right) + C \end{align*}$$

Answer 6

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

Answer 7

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.