Q. \[ \int \cos^3(x)\,dx \]

Q: How do you integrate \cos^3(x) ?

Use \cos^3x=\cos x(1-\sin^2x). Let u=\sin x, so du=\cos x\,dx. Then \int \cos^3x\,dx=\int (1-u^2)\,du=u-\frac{u^3}{3}+C=\sin x-\frac{\sin^3 x}{3}+C.

Q: What substitution method works for \int \cos^3(x)\,dx ?

Take u=\sin x. Rewrite \cos^3x=\cos x\cdot \cos^2x=\cos x(1-\sin^2x). Then integrate \int (1-u^2)\,du=u-\frac{u^3}{3}+C. Substitute back to get \sin x-\frac{\sin^3x}{3}+C.

Q: Can we use the identity \cos^3x=\frac{1}{4}(3\cos x+\cos 3x) ?

Yes. If \cos^3x=\frac{1}{4}(3\cos x+\cos 3x), then \int \cos^3x\,dx=\frac{1}{4}\left(3\int \cos x\,dx+\int \cos 3x\,dx\right)=\frac{1}{4}\left(3\sin x+\frac{1}{3}\sin 3x\right)+C.

Q: Are the two answers equivalent?

They are. Using \sin 3x=3\sin x-4\sin^3x, substitute into \frac{1}{4}\left(3\sin x+\frac{1}{3}\sin 3x\right) to get \sin x-\frac{\sin^3x}{3}+C.

Q: What is \frac{d}{dx}\left(\sin x-\frac{\sin^3 x}{3}\right) ?

Differentiate: \frac{d}{dx}\left(\sin x\right)=\cos x and \frac{d}{dx}\left(\frac{\sin^3x}{3}\right)=\sin^2x\cos x. So the derivative is \cos x-\sin^2x\cos x=\cos x(1-\sin^2x)=\cos x\cos^2x=\cos^3x.

Q: How do you integrate \cos^3(x) using power-reduction formulas?

Convert using \cos^3x=\frac{1}{4}(3\cos x+\cos 3x). Then integrate termwise: \int \cos x\,dx=\sin x and \int \cos 3x\,dx=\frac{1}{3}\sin 3x. Result: \frac{1}{4}\left(3\sin x+\frac{1}{3}\sin 3x\right)+C.

Answer

To integrate \( \cos^3(x) \), rewrite it as \( \cos(x)\cos^2(x) \). Use the identity \( \cos^2(x)=1-\sin^2(x) \).

\[
\int \cos^3(x)\,dx=\int \cos(x)\left(1-\sin^2(x)\right)\,dx
\]
Let \(u=\sin(x)\), so \(du=\cos(x)\,dx\).
\[
\int \cos(x)\left(1-\sin^2(x)\right)\,dx=\int (1-u^2)\,du
\]
\[
\int (1-u^2)\,du=u-\frac{u^3}{3}+C
\]
Substitute back \(u=\sin(x)\).
\[
\int \cos^3(x)\,dx=\sin(x)-\frac{\sin^3(x)}{3}+C
\]

Final result: \( \sin(x)-\frac{\sin^3(x)}{3}+C \)

Detailed Explanation

We want to compute the indefinite integral

\[
\int \cos^3(x)\,dx.
\]

Step 1: Rewrite \(\cos^3(x)\) to use a substitution.

A common strategy is to separate one factor of \(\cos(x)\) and convert the remaining factors using \(\sin^2(x)=1-\cos^2(x)\).

Write

\[
\cos^3(x)=\cos(x)\cos^2(x).
\]

Now use the identity \(\cos^2(x)=1-\sin^2(x)\):

\[
\cos^3(x)=\cos(x)\left(1-\sin^2(x)\right).
\]

So the integral becomes

\[
\int \cos^3(x)\,dx=\int \cos(x)\left(1-\sin^2(x)\right)\,dx.
\]

Step 2: Substitute.

Let

\[
u=\sin(x).
\]

Then differentiate:

\[
du=\cos(x)\,dx.
\]

This substitution exactly matches the \(\cos(x)\,dx\) part of the integrand.

So we rewrite the integral in terms of \(u\):

\[
\int \cos(x)\left(1-\sin^2(x)\right)\,dx=\int \left(1-u^2\right)\,du.
\]

Step 3: Integrate the polynomial.

Now expand and integrate term-by-term:

\[
\int \left(1-u^2\right)\,du=\int 1\,du-\int u^2\,du.
\]

\[
= u-\frac{u^3}{3}+C.
\]

Step 4: Substitute back \(u=\sin(x)\).

\[
u-\frac{u^3}{3}+C=\sin(x)-\frac{\sin^3(x)}{3}+C.
\]

Final Answer

\[
\int \cos^3(x)\,dx=\sin(x)-\frac{\sin^3(x)}{3}+C.
\]

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Calculus FAQ

How do you integrate \( \cos^3(x) \)?

Use \( \cos^3x=\cos x(1-\sin^2x)\). Let \(u=\sin x\), so \(du=\cos x\,dx\). Then \(\int \cos^3x\,dx=\int (1-u^2)\,du=u-\frac{u^3}{3}+C=\sin x-\frac{\sin^3 x}{3}+C\).

What substitution method works for \( \int \cos^3(x)\,dx \)?

Take \(u=\sin x\). Rewrite \( \cos^3x=\cos x\cdot \cos^2x=\cos x(1-\sin^2x)\). Then integrate \( \int (1-u^2)\,du=u-\frac{u^3}{3}+C\). Substitute back to get \( \sin x-\frac{\sin^3x}{3}+C\).

Can we use the identity \( \cos^3x=\frac{1}{4}(3\cos x+\cos 3x)\) ?

Yes. If \( \cos^3x=\frac{1}{4}(3\cos x+\cos 3x)\), then \(\int \cos^3x\,dx=\frac{1}{4}\left(3\int \cos x\,dx+\int \cos 3x\,dx\right)=\frac{1}{4}\left(3\sin x+\frac{1}{3}\sin 3x\right)+C\).

Are the two answers equivalent?

They are. Using \(\sin 3x=3\sin x-4\sin^3x\), substitute into \(\frac{1}{4}\left(3\sin x+\frac{1}{3}\sin 3x\right)\) to get \( \sin x-\frac{\sin^3x}{3}+C\).

What is \( \frac{d}{dx}\left(\sin x-\frac{\sin^3 x}{3}\right)\) ?

Differentiate: \(\frac{d}{dx}\left(\sin x\right)=\cos x\) and \(\frac{d}{dx}\left(\frac{\sin^3x}{3}\right)=\sin^2x\cos x\). So the derivative is \(\cos x-\sin^2x\cos x=\cos x(1-\sin^2x)=\cos x\cos^2x=\cos^3x\).

How do you integrate \( \cos^3(x)\) using power-reduction formulas?

Convert using \( \cos^3x=\frac{1}{4}(3\cos x+\cos 3x)\). Then integrate termwise: \(\int \cos x\,dx=\sin x\) and \(\int \cos 3x\,dx=\frac{1}{3}\sin 3x\). Result: \(\frac{1}{4}\left(3\sin x+\frac{1}{3}\sin 3x\right)+C\).

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