Question

Trig substitution question:
Integrate sqrt(4^2 -x^2) from 0 to 2.

My work:
x = 4sinθ x^2 = 4^2 *sin^2 θ
sinθ = x/4
sqrt(4^2(1-sin^2 θ))) = sqrt(4^2 *cos^2 θ) =4cosθ

integral 4cosθ = 4sinθ
sub for sinθ.... x from 0 to 2. Unfortunately, this isn't right. How should I have done the problem?

Answer 1

@Hero @robtobey @campbell_st

Answer 2

\[\int_0^2\sqrt{4^2-x^2}\;dx\\
\text{letting }x=4\sin\theta \iff \theta=\arcsin\frac{x}{4},\\
dx=4\cos\theta\;d\theta,\\
\text{making your integral}\\
\int_0^2\sqrt{4^2-\left(4\sin\theta\right)^2}\;\left(4\cos\theta\;d\theta\right)\]
Since you've changes the limit of integration, you have to change the limits, too. The interval [0,2] corresponds with x. Now, you're integrating with respect to theta. The new limits will be
\[\text{Upper: } \theta=\arcsin\frac{2}{4}=\arcsin\frac{1}{2}=\frac{\pi}{6}\\
\text{Lower: }\theta = \arcsin\frac{0}{4}=0\]
With the new limits, the integral is now
\[\large \int_0^{\frac{\pi}{6}}\sqrt{4^2-\left(4\sin\theta\right)^2}\;\left(4\cos\theta\;d\theta\right)\\
\large 16\int_0^{\frac{\pi}{6}}\sqrt{1-\sin^2\theta}\;\cos\theta\;d\theta\]