# Math Help - Integral. #6

1. ## Integral. #6

Evaluate : $\displaystyle \int_0^1 sin(\alpha x) sin(\beta x) \, dx$

where $\displaystyle \alpha$ and $\displaystyle \beta$ are roots for the equation $\displaystyle 2x=tan(x)$.

Using the forumla $\displaystyle sin(A) sin(B)=\frac{1}{2} ( cos(A-B) - cos(A+B) )$ is not allowed.

2. Well using that identity is the most obvious and easy way to solve the problem... Why can't you use it?

3. because the problem say that

4. Use $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ then.

5. Originally Posted by TWiX
Evaluate : $\displaystyle \int_0^1 sin(\alpha x) sin(\beta x) \, dx$

where $\displaystyle \alpha$ and $\displaystyle \beta$ are roots for the equation $\displaystyle 2x=tan(x)$.

Using the forumla $\displaystyle sin(A) sin(B)=\frac{1}{2} ( cos(A-B) - cos(A+B) )$ is not allowed.

Use integration by parts twice.

$u = \sin(\alpha x) \implies du=\alpha \cos(\alpha x)dx \qaud dv=\sin(\beta x) \implies v=-\frac{1}{\beta}\cos(\beta x)$

$I= \int\sin(\alpha x) \sin(\beta x)dx$

$I=-\frac{1}{\beta}\sin(\alpha x)\cos(\beta x)+\frac{\alpha}{\beta }\int \cos(\alpha x)\cos(\beta x)dx$

$u = \cos(\alpha x) \implies du=-\alpha \sin(\alpha x)dx \qaud dv=\cos(\beta x) \implies v=\frac{1}{\beta}\sin(\beta x)$

$I=-\frac{1}{\beta}\sin(\alpha x)\cos(\beta x)+\frac{\alpha}{\beta }\left( \frac{1}{\beta}\sin(\beta x)\cos(\alpha x) + \frac{\alpha}{\beta}\int \sin(\alpha x) \sin(\beta x)dx\right)$

This gives

$\left(1 -\frac{\alpha^2}{\beta^2} \right)I= -\frac{1}{\beta}\sin(\alpha x)\cos(\beta x) +\frac{\alpha}{\beta^2}\sin(\beta x)\cos(\alpha x)$

$-\beta \left(1 -\frac{\alpha^2}{\beta^2} \right)I=\sin(\alpha x)\cos(\beta x)\left[1-\frac{\alpha}{\beta}\frac{\tan(\beta x)}{\tan(\alpha x)} \right]$

Now if you evaluate for 0 to 1 the last factor gives

$1-\frac{\alpha}{\beta}\frac{\tan(\beta x)}{\tan(\alpha x)}]$

Since alpha and beta satify the equation

$2x=\tan(x)$ we get

$1-\frac{\alpha}{\beta}\frac{2\beta}{2\alpha}=0$