4
$\begingroup$

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

$\endgroup$
  • 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
4
$\begingroup$

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:

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.