How do you integrate: $$\int_{0}^{1}\frac{\sin(x)}{\sin(1-x)+\sin(x)}dx$$
The answer that you must get is $\frac{1}{2}.$
Asked
Active
Viewed 66 times
1
-
1Now let $1-x = y $ to get $I=-\int_{1}^{0} \frac{\sin(1-y)}{\sin(1-y)+\sin(y)} dy$ – King Tut Mar 01 '18 at 11:57
-
1Please, use MathJax. In the present form, the integral doesn't exist, though it's hard to tell if you just forgot a pair of parentheses. – Mar 01 '18 at 11:58
-
Oh yes I assumed two terms in denominator, only then we get that result.. – King Tut Mar 01 '18 at 12:00
-
Set $y=\dfrac{\pi x}2$ and use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx/439856#439856 – lab bhattacharjee Mar 01 '18 at 12:17
-
@Lab why do you need that pi/2 ? You can directly use your answer! – King Tut Mar 01 '18 at 12:27
-
@KingTut, With that transformation, the integrals become identical, right? – lab bhattacharjee Mar 01 '18 at 12:28
-
I was saying whats the harm in directly applying your answer's method. well i dont mind.. – King Tut Mar 01 '18 at 12:30
1 Answers
3
Let $I=\int_0^1\frac{\sin x}{\sin x+\sin (1-x)}\,dx~~~~~~~~~~~~~~~~(1)$
Take $x=1-z$. Then $I=\int_0^1\frac{\sin (1-z)}{\sin (1-z)+\sin (z)}\,dz$. Or we can write it as $$I=\int_0^1\frac{\sin (1-x)}{\sin x+\sin (1-x)}\,dx~~~~~~~~~(2)$$
Add $(1)$ and $(2)$, we have
$$2I=\int_0^1 1\,dx \implies I=\frac{1}{2}.$$
Goldy
- 611