UN---Integratable..

There is a much shorter way to this integral. We have :

$I=\int { \frac { dx }{ { sin }^{ 3 }x+{ cos }^{ 3 }x } } =\int { \frac { dx }{ (sinx+cosx)(1-sinxcosx) } }$

$\Rightarrow I=\int { \frac { (sinx+cosx)dx }{ (1+sin2x)(1-\frac { sin2x }{ 2 } ) } }$

Note : I have used ${(sinx+cosx)}^{2}=1+sin2x$ and $2sinxcosx=sin2x$

Put $sinx-cosx=t,(sinx+cosx)dx=dt$

Also ${t}^{2}=1-sin2x$ hence $sin(2x)=1-{t}^{2}$

$\Rightarrow I=2\int { \frac { dt }{ (2-{ t }^{ 2 })(1+{ t }^{ 2 }) } } =\frac { 2 }{ 3 } (\int { \frac { dt }{ 1+{ t }^{ 2 } } } +\int { \frac { dt }{ 2-{ t }^{ 2 } } } )$

$\Longrightarrow I=\frac { 2 }{ 3 } (\int { \frac { dt }{ 1+{ t }^{ 2 } } } +\frac { 1 }{ 2\sqrt { 2 } } (\int { \frac { dt }{ \sqrt { 2 } -t } } +\int { \frac { dt }{ \sqrt { 2 } +t } } ))$

Finally integrating we have :

$I=\frac { 2 }{ 3 } { tan }^{ -1 }t+\frac { 1 }{ 3\sqrt { 2 } } ln(\sqrt { 2 } +t)-\frac { 1 }{ 3\sqrt { 2 } } ln(\sqrt { 2 } -t)$

$\Rightarrow I=\frac { 2 }{ 3 } { tan }^{ -1 }t+\frac { 1 }{ 3\sqrt { 2 } } ln(\frac { \sqrt { 2 } +t }{ \sqrt { 2 } -t } )$

Putting back $t=sinx-cosx$ we have :

$I=\frac { 2 }{ 3 } { tan }^{ -1 }(sinx-cosx)+\frac { 1 }{ 3\sqrt { 2 } } ln(\frac { \sqrt { 2 } +sinx-cosx }{ \sqrt { 2 } -sinx+cosx } )$

Our integral can be written as :

$\displaystyle I=\int \mathrm{\frac{1}{(\sin x+\cos x)(1-\sin x\cos x)}}\mathrm{d}x$

Let $x=\frac{u}{2} , dx=\frac{1}{2} du$

$\displaystyle =>I=\frac{1}{2}\int \mathrm{\frac{1}{(\sqrt{\frac{1-\cos u}{2}}+\sqrt{\frac{1+\cos u}{2}})(1-\sqrt{\frac{1-\cos u}{2}}\sqrt{\frac{1+\cos u}{2}})}}\mathrm{d}u$

Rearranging and multiplying the numerator and the denominator by $\displaystyle \sqrt{\frac{1-\cos u}{2}}-\sqrt{\frac{1+\cos u}{2}}$ , we will end up with :

$\displaystyle I=\frac{1}{\sqrt{2}}\int \mathrm{\frac{\sqrt{1+\cos u}-\sqrt{1-\cos u}}{(\cos u)(2-\sin u)}}\mathrm{d}u$

But we have $\displaystyle \sqrt{1+\cos u}-\sqrt{1-\cos u}=\sqrt{(\sqrt{1+\cos u}-\sqrt{1-\cos u})^2}$

$\displaystyle=\sqrt{1+\cos u+1-\cos u -2\sqrt{(1-\cos u)(1+\cos u)}}=\sqrt{2-2\sin u}$

Hence, $\displaystyle I=\int \mathrm{\frac{\sqrt{1-\sin u}}{(\cos u)(2-\sin u)}}\mathrm{d}u$

$\displaystyle =\int \mathrm{\frac{\sqrt{1-\sin u}}{(\cos^2 u)(2-\sin u)}\cos u}\mathrm{d}u$

$\displaystyle = \int \mathrm{\frac{\sqrt{1-\sin u}}{(1-\sin u)(1+\sin u)(2-\sin u)}\cos u}\mathrm{d}u$

Let $\displaystyle y=\sin u , dy=\cos u du$

Therefore, $\displaystyle I=\int \mathrm{\frac{\sqrt{1-y}}{(1-y)(1+y)(2-y)}}\mathrm{d}y$

Now we use method of Partial fractions to get $\displaystyle \frac{1}{(1-y)(1+y)(2-y)}=\frac{1}{2} \frac{1}{1-y}+\frac{1}{6} \frac{1}{1+y} -\frac{1}{3} \frac{1}{2-y}$

and hence $\displaystyle I=\frac{1}{2} \int \mathrm{\frac{\sqrt{1-y}}{1-y}}\mathrm{d}y + \frac{1}{6} \int \mathrm{\frac{\sqrt{1-y}}{1+y}}\mathrm{d}y - \frac{1}{3} \int \mathrm{\frac{\sqrt{1-y}}{2-y}}\mathrm{d}y$

Now, we solve each integral alone:

$\displaystyle A= \int \mathrm{\frac{\sqrt{1-y}}{1-y}}\mathrm{d}y =\int \mathrm{(1-y)^{-\frac{1}{2}}}\mathrm{d}y = -2\sqrt{1-y}$

$\displaystyle B = \int \mathrm{\frac{\sqrt{1-y}}{1+y}}\mathrm{d}y$

Let $\displaystyle t=\sqrt{1-y} , y=1-t^2, dy=-2tdt$

$\displaystyle => B =2\int \mathrm{\frac{t^2}{t^2-2}}\mathrm{d}t = 2\int \mathrm{\frac{t^2-2}{t^2-2}+\frac{2}{t^2-2}}\mathrm{d}t= 2t+4\int \mathrm{\frac{1}{t^2-2}}\mathrm{d}t$

Using method of partial fractions, we get $\displaystyle \frac{1}{t^2-2}=\frac{1}{2\sqrt{2}} (\frac{1}{t-\sqrt{2}}-\frac{1}{t+\sqrt{2}})$ and hence:

$\displaystyle B= 2t+\sqrt{2}\int \mathrm{\frac{1}{t-\sqrt{2}}-\frac{1}{t+\sqrt{2}}}\mathrm{d}t=2t+\sqrt{2}(\ln(t-\sqrt{2})-\ln(t+\sqrt{2}))$

$\displaystyle = 2\sqrt{1-y}+\sqrt{2}(\ln(\sqrt{1-y}-\sqrt{2})-\ln(\sqrt{1-y}+\sqrt{2}))$

$\displaystyle C=\int \mathrm{\frac{\sqrt{1-y}}{2-y}}\mathrm{d}y$

Let $\displaystyle t=\sqrt{1-y}, y=1-t^2, dy=-2tdt$

$\displaystyle => C=-2\int \mathrm{\frac{t^2}{t^2+1}}\mathrm{d}t=-2\int \mathrm{\frac{t^2+1}{t^2+1}-\frac{1}{t^2+1}}\mathrm{d}t=-2t+2\arctan t$

$\displaystyle = -2\sqrt{1-y}+2\arctan\sqrt{1-y}$

Therefore,

$\displaystyle I=\frac{1}{2}A+\frac{1}{6}B-\frac{1}{3}C$ $\displaystyle = \frac{1}{2}(-2\sqrt{1-y})+\frac{1}{6}(2\sqrt{1-y}+\sqrt{2}(\ln(\sqrt{1-y}-\sqrt{2})-\ln(\sqrt{1-y}+\sqrt{2})))-\frac{1}{3}( -2\sqrt{1-y}+2\arctan\sqrt{1-y})$

Now, all what we have to do is substitute back $\displaystyle y=\sin u$ and $x=\frac{u}{2} <=> y=\sin 2x$ :

$=> \sqrt{1-y} = \sqrt{1-\sin 2x} = \sqrt{\cos^2x+\sin^2x-2\sin x \cos x}= \cos x-\sin x$

And by simplifying our answer , we end up with:

$\displaystyle \boxed{\int \mathrm{\frac{1}{(\sin^3 x+\cos^3 x)}}\mathrm{d}x= \frac{\sqrt{2}}{6}\ln(\frac{\cos x-\sin x -\sqrt{2}}{\cos x-\sin x +\sqrt{2}}) -\frac{2}{3}\arctan (\cos x-\sin x) +C}$

That's easy, it's $x$. Oh, that's cubes, not squares. I need new glasses.

Thanx to all.....really that's what I call a brilliant site.

Can you help me solve it

@Calvin Lin ,Sir can you help me in my post.

Sir Calvin I can't solve.....so can you help me out@Calvin Lin

Seems like Weierstrass substitution should give you a (potentially nasty) rational function, and those are all integrable by partial fractions.

Well, your problem is impossible to integrate. But, assuming you meant to integrate with respect to $x$, then

$\int \frac { 1 }{ \sin ^{ 3}{x } +\cos^{3}{x}} { dx }= -\frac { 2}{3 } (\tan ^{ -1 }{ (cos{x}-sin{x}) } +(1+i)(-1)^{3/4}\tanh ^{ -1 }{ (\frac { \tan ({ \frac { x }{ 2 })} -1 }{ \sqrt { 2 } } )} )+C$