7
$\begingroup$

$$\int_0^4 \frac{\left(y^2-4 y+5\right) \sin (y-2)}{2 y^2-8 y+11} d y$$ Here is my attempt $$ \begin{aligned} & y-2=t \Rightarrow d y=d t \\ & y^2-4 y+4=t^2 \Rightarrow 2 t^2=2 y^2-8 y+8 \\ & {y^2-4 y+5=t^2+1} \\ & 2 y^2-8 y+11=2 t^2+3 \\ & 2 y^2-8 y+10+1=2 t^3+3 \\ & 2\left(y^2-4 y+5\right)+1=2 t^3+3 . \\ & {\int_0^4 \frac{2\left(t^2+1\right) \sin t}{2\left(t^2+1\right)+1} d t} \\ & \Rightarrow \int_0^4\frac{\sin t}{\left.1+\frac{1}{2\left(t^2+1\right)}\right)} d t \end{aligned} $$

No other substitution seems to work. Any help or quickies on this one would be appreciated. Thanks

$\endgroup$

1 Answer 1

12
$\begingroup$

Observe that:

$$ \int_0^4 \frac{y^2-4y+5}{2y^2-8y+11}\,\sin(y-2)\,\text{d}y = \int_{-2}^2 \frac{x^2+1}{2x^2+3}\,\sin(x)\,\text{d}x = 0 $$

since the integration interval is centered at $O$ with respect to which the integrand is symmetric.

$\endgroup$
3
  • $\begingroup$ WOW! wait could you explain a little more in detail. I do know a property that the integral of an odd function is 0 if the limits of the integral are centered to O. But still I didnt get how you brought in x in the picture $\endgroup$ Mar 29, 2023 at 9:36
  • 1
    $\begingroup$ Aha Yes!!!. Beautiful Buddy $\endgroup$ Mar 29, 2023 at 9:39
  • $\begingroup$ So was @ElizabethHuffman's main mistake forgetting to change the integration extremes? $\endgroup$
    – Neil
    Mar 31, 2023 at 7:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .