Find the indefinite integral. (Use C for the constant of integration.)

Pis

Ask by Simmons May. in the United States

Mar 25,2025

Question

Find the indefinite integral. (Use C for the constant of integration.) $$ \int \sqrt{16-4 x^{2}} d x $$ Pis

UpStudy AI · Accepted Answer

We begin with the integral
$$
\int \sqrt{16-4x^2}\,dx.
$$
Notice that we can factor the expression under the square root:
$$
\sqrt{16-4x^2}=\sqrt{4(4-x^2)}=2\sqrt{4-x^2}.
$$
Thus the integral becomes
$$
\int \sqrt{16-4x^2}\,dx =\int 2\sqrt{4-x^2}\,dx.
$$

To evaluate the integral, we use the substitution
$$
x=2\sin\theta,\quad dx=2\cos\theta\,d\theta.
$$
Next, rewrite the square root:
$$
\sqrt{4-x^2}=\sqrt{4-4\sin^2\theta}=\sqrt{4(1-\sin^2\theta)}=2\cos\theta.
$$
Substitute these into the integral:
$$
\int 2\sqrt{4-x^2}\,dx =\int 2\cdot (2\cos\theta)\cdot (2\cos\theta\,d\theta)
=\int 8\cos^2\theta\,d\theta.
$$

We now simplify the integral using the double-angle identity for cosine:
$$
\cos^2\theta=\frac{1+\cos2\theta}{2}.
$$
Thus,
$$
8\cos^2\theta=8\cdot \frac{1+\cos2\theta}{2}=4(1+\cos2\theta).
$$
The integral becomes
$$
\int 8\cos^2\theta\,d\theta=4\int (1+\cos2\theta)\,d\theta.
$$
Integrate term by term:
$$
4\int 1\,d\theta=4\theta,\quad 4\int \cos2\theta\,d\theta=4\cdot\frac{1}{2}\sin2\theta=2\sin2\theta.
$$
Thus the result in terms of $$\theta$$ is
$$
4\theta+2\sin2\theta+C.
$$

Now we return to the original variable $$x$$. Recall the substitution:
$$
x=2\sin\theta\quad \Longrightarrow\quad \theta=\arcsin\frac{x}{2}.
$$
Also, using the double-angle formula,
$$
\sin2\theta=2\sin\theta\cos\theta.
$$
We already have
$$
\sin\theta=\frac{x}{2},
$$
and
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{x^2}{4}}=\frac{\sqrt{4-x^2}}{2}.
$$
Thus,
$$
\sin2\theta=2\cdot \frac{x}{2}\cdot \frac{\sqrt{4-x^2}}{2}
=\frac{x\sqrt{4-x^2}}{2}.
$$
Substitute back into the expression for the integral:
$$
4\theta+2\sin2\theta =4\arcsin\frac{x}{2}+2\left(\frac{x\sqrt{4-x^2}}{2}\right)
=4\arcsin\frac{x}{2}+x\sqrt{4-x^2}.
$$

Therefore, the indefinite integral is
$$
\boxed{4\arcsin\frac{x}{2}+x\sqrt{4-x^2}+C}.
$$
The indefinite integral is $$ 4\arcsin\frac{x}{2} + x\sqrt{4-x^2} + C $$.

Find the indefinite integral. (Use C for the constant of integration.)

Pis

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Find the indefinite integral. (Use C for the constant of integration.) Pis