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I am solving the following question

$$\int\frac{\sin x}{\sin^{3}x + \cos^{3}x}dx.$$

I have been able to reduce it to the following form by diving numerator and denominator by $\cos^{3}x$ and then substituting $\tan x$ for $t$ and am getting the following equation. Should Iis there any other way use partial fraction to integrate it further or

$$\int\frac{t}{t^3 + 1}dt.$$

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    Factor $t^3+1$ and calculate the partial fraction decomposition – 2017-02-28

4 Answers 4

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Hint:

$\int \frac{t}{t^3+1} dx = \int \frac{t+t^2-t^2}{t^3+1} dx =\int \frac{t(t+1)}{t^3+1} dx-\frac{1}{3}\int \frac{3t^2}{t^3+1} dx=\int \frac{t(t+1)}{(t+1)(t^2-t+1)} dx-\frac{1}{3}\ln(t^3+1)$

$=\frac{1}{2}\int \frac{2t}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)=\frac{1}{2}\int \frac{2t-1+1}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)$

$=\frac{1}{2}\int \frac{2t-1}{t^2-t+1} dx+\frac{1}{2}\int \frac{1}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)$

$=\frac{1}{2}\ln(t^2-t+1)+\frac{1}{2}\int \frac{1}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)$

The integral can be evaluated by partial fractions with complex roots.

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    This is the best answer, it's an important trick to split the fractions further, such that the derivative of the denominator is a multiple of the numerator – 2017-02-28
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Hint: $\frac{t}{t^{3}+1} = \frac{A}{t+1} + \frac{Bt + C}{t^{2}-t+1}$ where $A$ and $B$ and $C$ are constants to be found

Can you solve it now?

In case you get stuck:

$A = -\frac{1}{3}$ $B = \frac{1}{3}$ $C = \frac{1}{3}$

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

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    Should be $Bx+C$ for the last fraction – 2017-02-28
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    Id say that instead $\;B\;$ it should be $\;Bx+C\;$ ...unless you already did the calculations and found out $\;B=0\;$ ... – 2017-02-28
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    Yes realised that after actually working out what the fractions were, fixed now – 2017-02-28
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$\textbf{Hint.}$ Firstly,

$$f(t)=\frac{t}{t^3+1}=\frac{t}{(t+1)(t^2-t+1)}=\frac{t+1}{3(t^2-t+1)}-\frac{1}{3(t+1)}$$

The second integral is immediate (a logarithmic function). The first needs more work, but it can be reduced to the integral of a logaritmic plus an $\arctan$

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We have that $-1,\omega,\omega^{-1}$ are the roots of $t^3+1$. In particular, $\frac{t}{t^3+1}$ can be represented as $$ \frac{t}{t^3+1} = \frac{A}{t+1}+\frac{B}{t-\omega}+\frac{C}{t-\omega^{-1}}$$ with $A+B+C=0$ and $$ A = \lim_{t\to -1}\frac{t(t+1)}{t^3+1} = \lim_{t\to -1}\frac{t}{t^2-t+1} = -\frac{1}{3} $$ hence: $$ \frac{t}{t^3+1} = -\frac{1}{3}\cdot \frac{1}{t+1}+\frac{1}{3}\cdot\frac{t+1}{t^2-t+1} $$ and

$$ \int\frac{t}{t^3+1}\,dt = C+\frac{1}{\sqrt{3}}\,\arctan\left(\frac{2t-1}{\sqrt{3}}\right)+\frac{1}{3}\,\log(1+t)-\frac{1}{6}\,\log(1-t+t^2). $$