How Do The Integration of The Given Expression Is $27\pi$ Basically, this is a double integration question from the book Calculus 2 by Howard Anton. Half of the question is resolved and now I'm at this stage where I can't prove the left hand side to the right hand side.
$$
\int_{-3}^3 (6\sqrt {9-x^2} - 2x \sqrt{9-x^2}) dx = 27\pi
$$
 A: The thing is, you don't need to use substitutions or take integration of basic trigonometric functions.
All you need is to see $x*\sqrt{9-x^2}$ is odd function and have no contribution under integration over {-3, 3}. And for the rest, $6*\sqrt{9-x^2}$, you should see that it is 6 times the area of upper half of the circular region defined with center origin and radius 3, that is $ 6* \frac{9*{\pi} }{2}$
A: Substitute $x=3\sin u$ and use $\cos^2+\sin^2=1$. Note that the integral over the second term vanishes by symmetry.
Alternatively, note that the integral over the first term is the area of a semicircle.
A: Observe that $\sqrt{9 - x^2}$ is an even function and $x \sqrt{9 - x^2}$ is an odd function. Therefore, the given integral is equal to
\begin{equation*}
2 \times \int_0^3 6 \sqrt{9 - x^2} \,\mathrm dx = 12 \int_0^3 \sqrt{9 - x^2} \,\mathrm dx.
\end{equation*}
Now, substituting $x = 3 \sin t$, the integral becomes
\begin{align*}
12 \int_0^{\pi/2} (3 \cos t) (3 \cos t)\,\mathrm dt & = 12 \times 9 \int_0^{\pi/2} \cos^2 t\,\mathrm dt\\
& = 12 \times 9 \times \dfrac 1 2 \times \dfrac {\pi} 2\\
& = 27\pi.
\end{align*}
