`intsqrt(16-x^2) dx`

This question is in the form of the first substitution suggestion in this section, that is,

For `sqrt(a^2-x^2)`, use ` x =a sin theta`

So we have `a=4`, `x= 4 sin θ`, and `dx = 4 cos θ\ dθ`.

Substituting and simplifying the square root part first:

`sqrt(16-x^2) =sqrt(16-16 sin^2 theta)`

`=sqrt(16(1-sin^2 theta))`

`=4sqrt(cos^2 theta)`

`=4 cos theta`

Substituting into the integral gives:

`intsqrt(16-x^2) dx =int4 cos theta(4 cos theta\ d theta) =int16 cos^2 theta\ d theta`

`=16int1/2(cos 2 theta+1)d theta`

`=8(1/2sin 2 theta+theta)+K`

`=8(sin theta cos theta + theta)+K`

`=8(x/4(sqrt(16-x^2))/4+arcsin {:x/4:})+K`

`=(xsqrt(16-x^2))/2+8\ arcsin{:x/4:}+K`

The second-last step comes from drawing a triangle, like we did in earlier examples..

Quite often we can get different forms of the same final answer! That is, math software (or another human) can produce an answer which is actually correct, but different to the one given here.

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