How do you do this lol (1 Viewer)

idkkdi · May 24, 2022

Epicman69 said:
View attachment 35656

cursed

notme123 · May 25, 2022

Epicman69 said:
View attachment 35656

Let $y=\sin(x+\sin(x+\sin(x+...)))$
$x+y=x+\sin(x+\sin(x+\sin(x+...)))$
$\sin(x+y)=\sin(x+\sin(x+\sin(x+\sin(x+...))))=y$
$x+y=\arcsin(y)$
$x=\arcsin(y)-y$
We've got the inverse function of the integrand here. It's easy to prove using graphs but if you have a continuous function $f(x)$ and inverse function $g(y)$ , with points $(a,c)$ and $(b,d)$ , then $\int_a^b f(x) dx=bd-ac-\int_c^d g(y) dy.$ The relevant points on this curve are $(0,0)$ and $(\frac{\pi-2}{2},1)$ .So we can rewrite the integral as:
$\int_0^{\frac{\pi-2}{2}}\sin(x+\sin(x+\sin(x+...))) dx$
$= (\frac{\pi-2}{2})(1)-(0)(0)-\int_0^1(\arcsin(y)-y)dy$
$=\frac{\pi-2}{2}-{[y\arcsin(y)+\sqrt{1-y^2}-\frac{1}{2}y^2]}_0^1$
$=\frac{\pi-2}{2}-(\frac{\pi}{2}-\frac{1}{2}-1)$
$=\frac{1}{2}.$
I highly doubt this would ever be asked in a 4u test.

idkkdi · May 25, 2022

notme123 said:
Let $y=\sin(x+\sin(x+\sin(x+...)))$
$x+y=x+\sin(x+\sin(x+\sin(x+...)))$
$\sin(x+y)=\sin(x+\sin(x+\sin(x+\sin(x+...))))=y$
$x+y=\arcsin(y)$
$x=\arcsin(y)-y$
We've got the inverse function of the integrand here. It's easy to prove using graphs but if you have a continuous function $f(x)$ and inverse function $g(y)$ , with points $(a,c)$ and $(b,d)$ , then $\int_a^b f(x) dx=bd-ac-\int_c^d g(y) dy.$ The relevant points on this curve are $(\frac{\pi-2}{2},1)$ and $(0,0)$ .So we can rewrite the integral as:
$\int_0^{\frac{\pi-2}{2}}\sin(x+\sin(x+\sin(x+...))) dx$
$= (\frac{\pi-2}{2})(1)-0-\int_0^1(\arcsin(y)-y)dy$
$=\frac{\pi-2}{2}-{[y\arcsin(y)+\sqrt{1-y^2}-\frac{1}{2}y^2]}_0^1$
$=\frac{\pi-2}{2}-(\frac{\pi}{2}-\frac{1}{2}-1)$
$=\frac{1}{2}.$
I highly doubt this would ever be asked in a 4u test.

ye in a 4u test definitely would have parts at the least.

How do you do this lol (1 Viewer)

Epicman69

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idkkdi

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notme123

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idkkdi

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