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

Q: What is the integral \int \sin^3(x)\,dx ?

Write \sin^3(x)=\sin(x)(1-\cos^2(x)). Let u=\cos(x), du=-\sin(x)\,dx. Then \int \sin^3(x)\,dx = \cos^3(x)/3-\cos(x)+C.

Q: How do you choose the substitution for \int \sin(x)(1-\cos^2(x))\,dx ?

Use u=\cos(x) because du=-\sin(x)\,dx matches the factor \sin(x)\,dx. The integral becomes -\int (1-u^2)\,du.

Q: Can you compute \int \sin^3(x)\,dx using trig identities instead?

Use \sin^3(x)=\frac{3\sin(x)-\sin(3x)}{4}. Then integrate termwise: \int \sin^3(x)\,dx=\frac{1}{4}(-3\cos(x)+\frac{\cos(3x)}{3})+C.

Q: Are those two answers equivalent?

Yes. \cos^3(x)/3-\cos(x) = \frac{1}{4}\left(-3\cos(x)+\frac{\cos(3x)}{3}\right)+C using \cos(3x)=4\cos^3(x)-3\cos(x).

Q: What’s the derivative of \cos^3(x)/3-\cos(x) ?

Differentiate: \frac{d}{dx}\left(\frac{\cos^3(x)}{3}\right)=\cos^2(x)(-\sin(x)). Also \frac{d}{dx}(-\cos(x))=\sin(x). Sum gives \sin(x)(1-\cos^2(x))=\sin^3(x).

Q: What is the definite integral \int_{0}^{\pi} \sin^3(x)\,dx ?

Use symmetry: \int_{0}^{\pi}\sin^3(x)\,dx=2\int_{0}^{\pi/2}\sin^3(x)\,dx. Known result \int_{0}^{\pi/2}\sin^3(x)\,dx=2/3. So value =4/3.

Answer

Compute

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

Let \(u=\cos(x)\), then \(du=-\sin(x)\,dx\). So

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

Substitute back \(u=\cos(x)\):

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

Detailed Explanation

We want to compute the indefinite integral

\[
\int \sin^3(x)\,dx
\]

Step 1: Rewrite \( \sin^3(x) \) to use a standard identity.

A common approach is to separate one factor of \(\sin(x)\) and combine the other two using \( \sin^2(x) = 1 – \cos^2(x) \).

\[
\sin^3(x) = \sin(x)\sin^2(x)
\]

Step 2: Substitute the Pythagorean identity \( \sin^2(x) = 1 – \cos^2(x) \).

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

Now rewrite the integral:

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

Step 3: Use substitution.

Let

\[
u = \cos(x)
\]

Then differentiate both sides.

\[
du = -\sin(x)\,dx
\]

This means

\[
\sin(x)\,dx = -du
\]

Step 4: Rewrite the integral in terms of \(u\).

When \(u = \cos(x)\), we also have \(\cos^2(x) = u^2\).

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

Pull out the negative sign:

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

Step 5: Distribute and integrate term-by-term.

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

Integrate each part:

\[
-\int 1\,du = -u
\]
\[
\int u^2\,du = \frac{u^3}{3}
\]

So the result is:

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

Step 6: Substitute back \(u = \cos(x)\).

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

Final Answer:

\[
\int \sin^3(x)\,dx = -\cos(x) + \frac{1}{3}\cos^3(x) + C
\]

See full solution

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Calculus FAQ

What is the integral \( \int \sin^3(x)\,dx \) ?

Write \( \sin^3(x)=\sin(x)(1-\cos^2(x))\). Let \(u=\cos(x)\), \(du=-\sin(x)\,dx\). Then \( \int \sin^3(x)\,dx = \cos^3(x)/3-\cos(x)+C\).

How do you choose the substitution for \( \int \sin(x)(1-\cos^2(x))\,dx \) ?

Use \(u=\cos(x)\) because \(du=-\sin(x)\,dx\) matches the factor \(\sin(x)\,dx\). The integral becomes \(-\int (1-u^2)\,du\).

Can you compute \( \int \sin^3(x)\,dx \) using trig identities instead?

Use \( \sin^3(x)=\frac{3\sin(x)-\sin(3x)}{4}\). Then integrate termwise: \( \int \sin^3(x)\,dx=\frac{1}{4}(-3\cos(x)+\frac{\cos(3x)}{3})+C\).

Are those two answers equivalent?

Yes. \( \cos^3(x)/3-\cos(x) = \frac{1}{4}\left(-3\cos(x)+\frac{\cos(3x)}{3}\right)+C\) using \( \cos(3x)=4\cos^3(x)-3\cos(x)\).

What’s the derivative of \( \cos^3(x)/3-\cos(x) \) ?

Differentiate: \( \frac{d}{dx}\left(\frac{\cos^3(x)}{3}\right)=\cos^2(x)(-\sin(x))\). Also \( \frac{d}{dx}(-\cos(x))=\sin(x)\). Sum gives \( \sin(x)(1-\cos^2(x))=\sin^3(x)\).

What is the definite integral \( \int_{0}^{\pi} \sin^3(x)\,dx \) ?

Use symmetry: \(\int_{0}^{\pi}\sin^3(x)\,dx=2\int_{0}^{\pi/2}\sin^3(x)\,dx\). Known result \( \int_{0}^{\pi/2}\sin^3(x)\,dx=2/3\). So value \(=4/3\).

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