3
$\begingroup$

I am trying to integrate the function $$\int_{-\pi/2}^0 \sin(2x)\cos(nx) \, \mathrm{d} x.$$

My professor has an answer of $$\frac{-2\cos(\frac{n \pi}{2})+1}{n^{2}-4}.$$

When I do this problem, I first notice that it can be broken into its trig identity and the new integral is written as $$\frac{1}{2}\int_{-\pi/2}^0 \sin((2+n)x \, \mathrm{d} x + \frac{1}{2}\int_{-\pi/2}^0 \sin((2-n)x \, \mathrm{d} x$$

Once I integrate both of these, I end up getting $$\frac{-1}{2+n}\left[1-\cos\left(\frac{(2+n)\pi}{2}\right)\right]+\frac{-1}{2-n}\left[1-\cos\left(\frac{(2-n)\pi}{2}\right)\right]$$

I can't seem to manipulate the above equation to get what my professor got

$\endgroup$
2
$\begingroup$

$$\cos\left(\frac{(2\pm n)\pi}2\right)=\cos(\pi\pm \frac{n\pi}2)=-\cos\frac{n\pi}2$$ as $\cos(\pi\pm y)=\cos y\cos\pi\mp\sin y\sin\pi=-\cos y$ using $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ and $\cos\pi=-1,\sin\pi=0$

$\endgroup$
1
$\begingroup$

The correct answer is $$2\frac{1+\cos(\frac{n \pi}{2})}{n^{2}-4}.$$ After integrate you should get $$\dfrac12\left\{\frac{-1}{2+n}\left[1-\cos\left(\frac{(2+n)\pi}{2}\right)\right]+\frac{-1}{2-n}\left[1-\cos\left(\frac{(2-n)\pi}{2}\right)\right]\right\}.$$ The rest is in lab bhattacharjee's answer.

$\endgroup$

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.