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Evaluation of $\displaystyle \int\frac{1}{\sin^3 x-\cos^3 x}dx$

$\bf{My\; Try::}$ Given $\displaystyle \int\frac{1}{\sin^3 x-\cos^3 x}dx = \int\frac{1}{(\sin x-\cos x)\cdot (\sin^2 x-\sin x\cos x+\cos^2 x)}dx$

$\displaystyle = 2\int\frac{(\sin x-\cos x)}{(\sin x-\cos x)^2\cdot (2-\sin 2x)}dx = 2\int \frac{(\sin x-\cos x)}{(1-\sin 2x)\cdot (2-\sin 2x)}dx$

Let $(\sin x+\cos x) = t\;,$ Then $(\cos x -\sin x)dx = dt\Rightarrow (\sin x -\cos x)dx = dt$

and $(1+\sin 2x)=t^2\Rightarrow \sin 2x = (t^2-1)$

So Integral Convert into $\displaystyle 2\int\frac{1}{(2-t^2)\cdot (3-t^2)}dt = 2\int\frac{1}{(t^2-2)\cdot (t^2-3)}dt$

My Question is , is there is any better method other then that

Help me

Thanks.

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    $\begingroup$ I think initially your method is pretty cool! I would have factored the denom in the same way. But I suspect there is a negative sign missing when you go from $cosx-sinx$ to $sinx-cosx$ in front of that $dx$ term? $\endgroup$ – imranfat Feb 12 '14 at 17:14
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    $\begingroup$ I guess your method is the simplest one. For example, Weierstrass is cumbersome, $t = \sin\left(x\right)$ is quite bad, etc... $\endgroup$ – Felix Marin Feb 12 '14 at 17:41
  • $\begingroup$ Thanks imranfat and Felix Marin.. $\endgroup$ – juantheron Feb 15 '14 at 8:33
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You're missing a minus sign at one point, but other than that I think you're OK. Next, use partial fractions: $$ \frac{1}{(2-t^2)(3-t^2)} = \frac{A}{\sqrt{2}-t} + \frac{B}{\sqrt{2}+t} + \frac{C}{\sqrt{3}-t} + \frac{D}{\sqrt{3}+t} $$

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