To find the integral
[tex]\[
\int \frac{x+7}{\sqrt{16-(x-4)^2}} \, dx,
\][/tex]
we need to evaluate it step-by-step. Here is the detailed solution:
1. Recognize the Structure of the Integrand:
The presence of [tex]\(\sqrt{16 - (x-4)^2}\)[/tex] suggests that we might be dealing with a form conducive to trigonometric substitution.
2. Rewriting the Integrand with a Suitable Substitution:
Let’s consider a substitution that simplifies [tex]\(\sqrt{16 - (x-4)^2}\)[/tex]. We can use [tex]\( u = x - 4 \)[/tex], thus [tex]\( du = dx\)[/tex], and the integral becomes:
[tex]\[
\int \frac{u+11}{\sqrt{16-u^2}} \, du,
\][/tex]
where [tex]\(u = x - 4\)[/tex] and [tex]\(x = u + 4\)[/tex], so:
[tex]\[
(x + 7) = (u + 4 + 7) = (u + 11).
\][/tex]
3. Simplify Using Trigonometric Substitution:
To deal with [tex]\(\sqrt{16 - u^2}\)[/tex], use the substitution [tex]\(u = 4 \sin \theta\)[/tex], [tex]\(\sqrt{16 - u^2} = \sqrt{16 - 16\sin^2 \theta} = 4 \cos \theta\)[/tex], and [tex]\(du = 4 \cos \theta \, d\theta\)[/tex].
Substitute these into the integral:
[tex]\[
\int \frac{4 \sin \theta + 11}{4 \cos \theta} \cdot 4 \cos \theta \, d\theta = \int (4 \sin \theta + 11) \, d\theta.
\][/tex]
4. Separate the Integral:
[tex]\[
\int (4 \sin \theta + 11) \, d\theta = 4 \int \sin \theta \, d\theta + 11 \int \, d\theta
\][/tex]
5. Integrate Each Term:
[tex]\[
4 \int \sin \theta \, d\theta = -4 \cos \theta + C_1,
\][/tex]
[tex]\[
11 \int \, d\theta = 11\theta + C_2.
\][/tex]
Combining these, we have:
[tex]\[
-4 \cos \theta + 11 \theta + C.
\][/tex]
6. Return to Original Variable:
Recall [tex]\( u = 4 \sin \theta \)[/tex], thus [tex]\(\sin \theta = \frac{u}{4}\)[/tex] and [tex]\(\theta = \arcsin \left( \frac{u}{4} \right)\)[/tex]:
[tex]\[
\cos \theta = \sqrt{1 - \sin^2 \theta}=\sqrt{1 - \left(\frac{u}{4}\right)^2}=\sqrt{\frac{16 - u^2}{16}} = \frac{\sqrt{16 - u^2}}{4}.
\][/tex]
Substitute back [tex]\(u = x - 4\)[/tex]:
[tex]\[
\cos \theta = \frac{\sqrt{16 - (x-4)^2}}{4}.
\][/tex]
So, the integral becomes:
[tex]\[
-\sqrt{16 - (x-4)^2} + 11 \arcsin \left( \frac{x-4}{4} \right) + C.
\][/tex]
With some simplification:
[tex]\[
-\sqrt{22x - (x+7)^2 + 49} + 11 \arcsin \left( \frac{x-4}{4} \right) + C.
\][/tex]
Therefore, the integral of [tex]\(\frac{x+7}{\sqrt{16-(x-4)^2}} \, dx\)[/tex] is:
[tex]\[
-\sqrt{22x - (x+7)^2 + 49} + 11 \arcsin \left( \frac{x-4}{4} \right) + C.
\][/tex]
