Evaluation of $\displaystyle \int \sin (2015x)\cdot \sin^{2013}(x)dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int \sin (2015x)\cdot \sin^{2013}(x)dx = \int \sin (2014x+x)\cdot \sin^{2013}(x)dx$

So $\displaystyle I = \int \left(\sin 2014x\cdot \cos x+\cos 2014x\cdot \sin x\right)\cdot \sin^{2013}(x)dx $

So $\displaystyle I = \int \sin (2014x)\cdot \sin^{2013}(x)\cdot \cos x dx +\int \cos (2014x)\cdot \sin^{2014}(x)dx$

Now Using Integration by parts for $\bf{(I)}$ Integral..

So $\displaystyle I = \sin (2014x)\cdot \frac{\sin^{2014}(x)}{2014}-\frac{2014}{2014}\int \cos(2014x)\cdot \sin^{2014}(x)dx+\int \cos (2014x)\cdot \sin^{2014}(x)dx$

So $\displaystyle I = \sin (2014 x)\cdot \frac{\sin^{2014}(x)}{2014}+\mathcal {C}$

Can We solve it using Complex no. Something like $\displaystyle \sin x = \frac{e^{ix}-e^{-ix}}{2i}$

If yes, Then plz explain it to me , Thanks

  • 1
    $\begingroup$ You've solved it with a simple method but why would you take another hard way? $\endgroup$ – Venus Oct 7 '14 at 15:27

Observe that $\sin{2015x} \equiv \Im{(z^{2015})}$

Where $z = \cos{x}+ i \sin{x}$

If $z = \cos{x} + i\sin{x}$, then $\text{d}z = (i\cos{x} - \sin{x})\text{d}x = i(\cos{x} + i\sin{x})\text{d}x = iz\text{d}x$

So the integral in question, is this:

$$\int \Im{\left[z^{2015}\left(\frac{z^2-1}{2iz}\right)^{2013}\frac{\text{d}z}{iz}\right]} = \Im{\left[\int \left(\frac{z^2-1}{2i}\right)^{2013} \text{d}\left(\frac{z^2-1}{2i}\right)\right]}$$

$$= \Im{\left[ \frac{1}{2014}\left(\frac{z^2-1}{2i}\right)^{2014} \right]} = \frac{1}{2014}\Im{\left[ \left(\frac{z-z^{-1}}{2i}\right)^{2014} z^{2014} \right]}$$

$$= \frac{1}{2014} \Im{\left[\sin^{2014}{x} z^{2014}\right]} = \frac{1}{2014}\sin{2014x}\sin^{2014}{x} +\mathcal{C}$$


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