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$$I = \int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}}$$

Thought for a while but cannot seem to find how do i procced with this integral.

Another variant of this problem is $$I = \int \frac{\sin^{3}x+\cos^{3}x}{\sin^{2}{x}\cos^{2}{x}}$$ where you just seperate the terms in numerator and it simplifies to $$I=\int \sec{x}\tan{x}dx+\int \csc{x}\cot{x}dx$$ which gives $$\sec{x}-\csc{x}+C$$ on integrating.

But in this case it doesn't simplify into something straightforward.

2 Answers 2

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Notice that \begin{eqnarray} \dfrac{\sin^3x+\cos^3x}{\sin x\cos x}&=& \dfrac{\sin^3x}{\sin x\cos x}+ \dfrac{\cos^3x}{\sin x\cos x}\\ &=&\dfrac{\sin^2x}{\cos x}+\dfrac{\cos^2x}{\sin x}\\ &=&\dfrac{1-\cos^2x}{\cos x}+\dfrac{1-\sin^2x}{\sin x}\\ &=&\dfrac{1}{\cos x}-\dfrac{\cos^2x}{\cos x}+\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}\\ &=&\dfrac{1}{\cos x}+\dfrac{1}{\sin x}-\cos x-\sin x \end{eqnarray} To compute the two integrals $$ \int\dfrac{1}{\cos x}\,dx \text{ and } \int\dfrac{1}{\sin x}\,dx $$ we use the substitution $$ t=\tan\dfrac{x}{2} \text{ or equivalently } x=2\tan^{-1}t. $$ Since \begin{eqnarray} \cos x&=&\dfrac{\cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}}=\dfrac{1-t^2}{1+t^2}\\ \sin x&=&\dfrac{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}}=\dfrac{2t}{1+t^2}\\ dx&=&\dfrac{2}{1+t^2}\,dt \end{eqnarray} we have \begin{eqnarray} \int\dfrac{1}{\cos x}\,dx&=&\int\dfrac{1+t^2}{1-t^2}\cdot\dfrac{2}{1+t^2}\,dt=\int\dfrac{2}{1-t^2}\,dt=\int\left(\dfrac{1}{1-t}+\dfrac{1}{1+t}\right)\,dt=\ln\left|\dfrac{1+t}{1-t}\right|+c_1\\ \int\dfrac{1}{\sin x}\,dx&=&\int\dfrac{1+t^2}{2t}\cdot\dfrac{2}{1+t^2}\,dt=\int\dfrac{1}{t}\,dt=\ln|t|+c_2 \end{eqnarray} Hence \begin{eqnarray} \int\dfrac{\sin^3x+\cos^3x}{\sin x\cos x}\,dx&=&\int\left(\dfrac{1}{\cos x}+\dfrac{1}{\sin x}-\cos x-\sin x\right)\,dx\\ &=&\ln\left|\dfrac{1+\tan\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}\right|+\ln\left|\tan\dfrac{x}{2}\right|-\sin x+\cos x+c \end{eqnarray}

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    ok i agree with you till where you simplified the expression to $$ \int \sec{x}dx-\int \cos{x}dx+\int \csc{x}dx-\int \sin{x}dx$$ $$\Rightarrow \log|\sec{x}+\tan{x}|-\sin{x}+\log|\csc{x}-\cot{x}|+\cos{x}+C$$ The substitution you did after that was unnecessary! – 2017-02-11
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    @mikealise Your statement doesn't really prove that my answer is wrong! Does it? – 2017-02-11
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    Didn't say your answer is wrong, in fact I upvoted it! – 2017-02-11
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\begin{align} \sin^3x+\cos^3x &=(\sin x+\cos x)(\sin^2x+\cos^2x-\sin x \cos x) \\ &=(\sin x+\cos )(1-\sin x \cos x) \end{align}

Then \begin{align} I &=\int \dfrac{(\sin x+\cos )(1-\sin x \cos x)}{\sin x \cos x}\,\mathrm{d}x \\ &=\int \dfrac{\sin x +\cos x}{\sin x \cos x}\,\mathrm{d}x-\int\,\mathrm{d}x \\ &=\int \sec x\,\mathrm{d}x+\int \csc x\,\mathrm{d}x-\int \,\mathrm{d}x \\ &=\ln \left| {\sec x+\tan x} \right|-\ln|\csc x+\cot x|-x+C \\ &=\ln \left| \dfrac{\sec x+\tan x}{\csc x+\cot x}\right|-x+C, \\ \end{align} where $C$ is an arbitrary constant.