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Recently I have seen Syamaprasad Chakrabarti's post and suddenly one question stuck in my mind.

For Syamaprasad's integral there is a closed form summation formula of $\frac{\sin(nx)}{\sin(x)}$. But suppose if I give an integral like

$$\int\frac{\tan(23x)}{\tan(x)}dx$$ then I don't think that we can express this integral as a closed form of summation.

Why? Because I know that we can find the summations of $\sin$ and $\cos$ series but not for $\tan$ series.

$\sin(a)+\sin(a+b)+\sin(a+2b)+\sin(a+3b)+...+\sin(a+(n-1)b)=\frac{\sin(\frac{nb}{2})}{\sin(\frac{b}{2})}(\sin(a+(n-1)\frac{b}{2})$

And

$\cos(a)+\cos(a+b)+\cos(a+2b)+\cos(a+3b)+...+\cos(a+(n-1)b)=\frac{\sin(\frac{nb}{2})}{\sin(\frac{b}{2})}(\cos(a+(n-1)\frac{b}{2})$

But There is no summation formula of

$\tan(a)+\tan(a+b)+\tan(a+2b)+...+\tan(a+(n-1)b)$.

Then how to integrate $$\int\frac{\tan(23x)}{\tan(x)}dx$$ ?

Please help me.

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Krishna Aggarwal
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1 Answers1

3

From

$$\cos(nx)+i\sin(nx)=(e^{ix})^n=(c+is)^n\\=\sum_{\text{even k}} (-1)^{k/2}\binom nk c^ks^{n-k}+i\sum_{\text{odd k}} (-1)^{(k-1)/2}\binom nk c^ks^{n-k}$$ you draw

$$\tan(nx)=\frac{\displaystyle\sum_{\text{odd k}} (-1)^{(k-1)/2}\binom nk c^ks^{n-k}}{\displaystyle\sum_{\text{even k}} (-1)^{k/2}\binom nk c^ks^{n-k}}=\frac{\displaystyle\sum_{\text{odd k}} (-1)^{(k-1)/2}\binom nk \tan^{n-k}(x)}{\displaystyle\sum_{\text{even k}} (-1)^{k/2}\binom nk \tan^{n-k}(x)}.$$

Then by the substitution $x=\arctan(t)$, the integrand becomes the rational fraction

$$\frac{\displaystyle\sum_{\text{odd k}} (-1)^{(k-1)/2}\binom nk t^{n-k}}{\displaystyle\sum_{\text{even k}} (-1)^{k/2}\binom nk t^{n-k}}\frac1{t\,(t^2+1)}.$$

The roots of the denominator are known, so you can decompose in elementary fractions. Roll up your sleaves.