Question

Let $\alpha ,\beta$ be the distincet positive roots of the equation $\tan x=2x$ then evaluate ${\int }_0^1(\sin \alpha x.\sin \beta x)dx$ independent of $\alpha ,\beta$.

• A. 0
• B. -2
• C. 1
• D. 2

Answer 1

The correct option is A 0

${\int }_0^1(\sin \alpha x.\sin \beta x)dx$
$=\frac{1}{2}{\int }_0^1\left(cos\right(\alpha -\beta )dx-cos(\alpha +\beta \left)x\right)dx$
$=\frac{1}{2}(\frac{sin(\alpha -\beta )x}{\alpha -\beta }-\frac{sin(\alpha +\beta )x}{\alpha +\beta }){|}_0^1$
$=\frac{1}{2}(\frac{sin(\alpha -\beta )}{\alpha -\beta }-\frac{sin(\alpha +\beta )}{\alpha +\beta })$
Now given $\alpha$ and $\beta$ are roots
so
$tan\alpha =2\alpha$ and
$tan\beta =2\beta$
Subsituting $\alpha -\beta$ and $\alpha +\beta$ in the above integral, gives
$I=0$