# Thread: 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.

Final Answer = 0

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.

Final Answer = 0

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

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