How do I integrate $\int \frac{dx}{\sin^3 x + \cos^3 x}$? [closed]

How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?

It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.

• Try breaking up the result using partial fractions Mar 12, 2016 at 16:15
• The terms are not linear in sin(x) or cos(x). I do not know how to decompose this Mar 12, 2016 at 16:18
• @AgnishomChattopadhyay $$\sin^3x+\cos^3x=(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=(\sin^3x+\cos^3x)\left(1+\frac12\sin2x\right)$$ Mar 12, 2016 at 16:49
• I still do not see it. Also, the first factor in the last time should probably be sin(x) + cos(x) Mar 12, 2016 at 16:53
• Aug 27, 2018 at 6:33

\begin{align} \sin^3(x)+\cos^3(x) & = (\sin(x)+\cos(x))(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)) \\ & \{\sin(x)+\cos(x) = \sqrt{2}\sin(x+\pi/4)\} \\ & = \sqrt{2}\sin(x+\pi/4)(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x)) \\ & \{\sin(2x) = 2\sin(x)\cos(x)\} \\ & =\sqrt{2}\sin(x+\pi /4)(\sin^2(x) + \cos^2(x) -\sin(2x)/2)) \\ & =\sqrt{2}\sin(x+\pi /4)(1-\sin(2x)/2)) \\ & \{u = x +\pi/4, \: \sin(2x)=-\cos(2u)\} \\ & =(\sqrt{2}/2) \sin(u)(2+\cos(2u)) \\ & \{\cos^2(u) = \frac{1}{2}(1 + \cos(2u)\} \\ & =(\sqrt{2}/2) \sin(u)(1+2 \cos^2(u)) \\ \end{align}

\begin{align} \ \int \frac{1}{\sin^3 (x) + \cos^3 (x)}dx & = \int{\frac{\sqrt{2}}{\sin(u)(1+2\cos^2(u))}du} \\ & =\int{\frac{\sqrt{2}\sin(u)}{\sin^2(u)(1+2\cos^2(u))}du} \\ & \{\cos(u)=t, \:-\sin(u) du = dt\} \\ & =-\int{\frac{\sqrt{2}}{(1-t^2)(1+2t^2)}dt} \\ & =-\sqrt{2} \int{\frac{1/6}{1-t}+\frac{1/6}{1+t}+\frac{2/3}{1+2t^2}dt} \\ \end{align}

Maybe this can help :

$$\frac{2}{\sin x+\cos x}+\frac{\sin x+\cos x}{1-\cos x\sin x}=\frac{3}{\sin^3 x+\cos ^3 x}$$ and also $$\sin x+\cos x=\sqrt 2 \sin \big( x+ \frac{\pi}{4} \big).$$

$\;\;\displaystyle\frac{1}{(\sin x+\cos x)(1-\sin x\cos x)}=A\left(\frac{\sin x+\cos x}{1-\sin x\cos x}\right)+B\left(\frac{1}{\sin x+\cos x}\right)$

where $1=A(\sin x+\cos x)^2+B(1-\sin x\cos x)=(A+B)+(2A-B)\sin x\cos x$.

Then $A=\frac{1}{3}$ and $B=\frac{2}{3}$,

so $\displaystyle\int\frac{1}{\sin^3 x+\cos^3 x}dx=\int\left(\frac{1}{3}\cdot\frac{\sin x+\cos x}{1-\sin x\cos x}+\frac{2}{3}\cdot\frac{1}{\sin x+\cos x}\right)dx$

$\displaystyle=\frac{1}{3}\int\frac{2\sin x+2\cos x}{2-2\sin x\cos x}dx+\frac{2}{3}\int\frac{1}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx$

$\displaystyle=\frac{1}{3}\int\frac{\sin x+\cos x}{1+(\sin x-\cos x)^2}dx+\frac{2}{3\sqrt{2}}\int\csc\left(x+\frac{\pi}{4}\right)dx$

$\displaystyle=\frac{1}{3}\arctan(\sin x-\cos x)+\frac{2}{3\sqrt{2}}\ln\big|\csc\left(x+\frac{\pi}{4}\right)-\cot\left(x+\frac{\pi}{4}\right)\big|+C$

$\displaystyle=\frac{1}{3}\arctan(\sin x-\cos x)+\frac{2}{3\sqrt{2}}\ln\left\vert\frac{\sqrt{2}-\cos x+\sin x}{\sin x+\cos x}\right\vert+C$

Maybe this will help:

$$=\int\frac{\tan^3x}{\tan^3x(\cos^3x+\sin^3x)}dx=\int\frac{\tan^3x}{\sin^3x(1+\tan^3x)}dx=\int\frac{\sec^3x}{1+\tan^3x}dx\\=\int\frac{(1+\tan^2 x)\sec x}{1+\tan^3x}dx$$

• I'm sorry; I'm still not seeing it. How does this help? Mar 12, 2016 at 18:07