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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:

$$intsqrt81-x^2 , dx$$

Answer key:

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

My answer:

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

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.

calculus integration trigonometric-integrals

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edited might 31 "17 in ~ 1:41

Matthew Conroy

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asked might 31 "17 at 1:24

L1ghtShadowL1ghtShadow

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edited might 31 "17 in ~ 1:41

Matthew Conroy

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asked might 31 "17 at 1:24

L1ghtShadowL1ghtShadow

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$endgroup$

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## 3 answers 3

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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.