What is $\int \frac{x + 5}{\sqrt{9 - {(x - 3)}^{2}}} d x$ $int (x+5)/sqrt(9-(x-3)^2) dx$ ?

Question

2124 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-22T13:17:32

See below

Lets make the change $x - 3 = 3 sin t$ $x-3=3sint$ with this change

$d x = 3 cos t d t$ $dx=3costdt$ and $x + 5 = 3 sin t + 8$ $x+5=3sint+8$

$I = \int \frac{3 sin t + 8}{\sqrt{3^{2} - 3^{2} {sin}^{2} t}} \cdot 3 cos t d t =$ $I=int(3sint+8)/sqrt(3^2-3^2sin^2t)·3costdt=$

$= \int \frac{(3 sin t + 8) (3 cos t d t)}{3 (\sqrt{1 - {sin}^{2} t}) =}$ $=int((3sint+8)(3costdt))/(3(sqrt(1-sin^2t))=$

$= \int 3 (sin t + 8) d t = - 3 cos t + 24 t + C =$ $=int3(sint+8)dt=-3cost+24t+C=$

With change made $sin t = \frac{x - 3}{3}$ $sint=(x-3)/3$ and $\sqrt{1 - \frac{{(x - 3)}^{2}}{9}} = cos t =$ $sqrt(1-(x-3)^2/9)=cost=$

$\frac{\sqrt{9 - x^{2} + 6 x - 9}}{3} = cos t$ $sqrt(9-x^2+6x-9)/3=cost$ and $t = arcsin (\frac{x - 3}{3})$ $t=arcsin((x-3)/3)$

Finally we have $I = - \sqrt{6 x - x^{2}} + 24 arcsin (\frac{x - 3}{3}) + C$ $I=-sqrt(6x-x^2)+24arcsin((x-3)/3)+C$

What is int (x+5)/sqrt(9-(x-3)^2) dx∫x+5√9−(x−3)2dxint (x+5)/sqrt(9-(x-3)^2) dx?