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So I recognize this problem needs trig below to solve and I placed it ~ above the next opposite $heta$ and also I obtained a various answer since from the an essential since they placed it on the length and also not the elevation of the triangle.

I obtained a very comparable answer, the only difference is that every little thing except C is - and also instead of arcsin I had arccos.

Integral:

$$intsqrt81-x^2 , dx$$

$$frac12xsqrt81-x^2+frac812arcsinfracx9+C$$

$$-frac12xsqrt81-x^2-frac812arccosfracx9+C$$

To define what I typical by opposite of $heta$. In my triangle, $cos heta=fracx9$. With the price key"s collection up, $cos heta=fracsqrt81-x^29$.

My concern is just how do you understand where to put the radical on her triangle? that seems prefer it shouldn"t matter due to the fact that the triangle works.

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edited might 31 "17 in ~ 1:41 Matthew Conroy
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$int sqrt 81-x^2 dx$ don"t forget the index of integration.

Either one of these substitutions is acceptable.

$x = 9cos t$ or $x = 9sin t$

Lets show both ways

$int sqrt 81-x^2 dx\x = 9cos t\dx = -9sin t\int sqrt 81-81cos^2 t(-9sin t) dt\int - 81 sin^2 t dt\int - frac 812 + frac 812cos 2t dt\ - frac 812t + frac 814sin 2t\- frac 812t + frac 812 sin tcos t+C\t = arccos frac x9\- frac 812arccos frac x9 + frac 812 (sqrt 1-frac x^281)( frac x9)+C\- frac 812arccos frac x9 + frac 12 xsqrt 81-x^2 + C$

$int sqrt 81-x^2 dx\x = 9sin t\dx = 9cos t\int sqrt 81-81sin^2 t(9cos t) dt\int 81 cos^2 t dt\int frac 812 + frac 812cos 2t dt\ frac 812t + frac 814sin 2t\frac 812t + frac 812 sin tcos t+C\t = arcsin frac x9\frac 812arcsin frac x9 + frac 812 frac x9sqrt 1-frac x^281 +C\frac 812arcsin frac x9 + frac 12 xsqrt 81-x^2 + C$

Both are fine ways to go.

$- frac 812arccos frac x9 + frac 12 xsqrt 81-x^2 + C = frac 812arcsin frac x9 + frac 12 xsqrt 81-x^2 + D$

as $-frac 812arccos frac x9$ differs from $frac 812arcsin frac x9$ by a constant.

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Generally, girlfriend will see the $x = a sin t$ substitution since it has actually fewer minus signs.