Q. \[ \int \cos^2(x)\,dx \]

Q: How do I integrate \cos^2(x) using a power-reduction identity?

Use \cos^2(x)=\frac{1+\cos(2x)}{2} . Then \[ \int \cos^2(x)\,dx=\frac{1}{2}\int 1\,dx+\frac{1}{2}\int \cos(2x)\,dx=\frac{x}{2}+\frac{\sin(2x)}{4}+C. \]

Q: What is a quick way using \sin(2x) form?

Since \sin(2x)=2\sin(x)\cos(x) , from the identity above you get \[ \int \cos^2(x)\,dx=\frac{x}{2}+\frac{\sin(2x)}{4}+C. \]

Q: Can I integrate \cos^2(x) by rewriting as 1-\sin^2(x) ?

Use \cos^2(x)=1-\sin^2(x) . Then \[ \int \cos^2(x)\,dx=\int 1\,dx-\int \sin^2(x)\,dx. \] With \int \sin^2(x)\,dx=\frac{x}{2}-\frac{\sin(2x)}{4}+C , the result is \frac{x}{2}+\frac{\sin(2x)}{4}+C .

Q: How do I check my answer by differentiating?

Differentiate \frac{x}{2}+\frac{\sin(2x)}{4} : \[ \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{1}{2},\quad \frac{d}{dx}\left(\frac{\sin(2x)}{4}\right)=\frac{1}{4}\cdot 2\cos(2x)=\frac{\cos(2x)}{2}. \] Sum gives \frac{1+\cos(2x)}{2}=\cos^2(x).

Q: What is the general method for \int \cos^n(x)\,dx when n is even?

For even powers, use identities like power-reduction for \cos^2(x) . Often you rewrite \cos^n(x) into sums of cosines of multiples of x (via half-angle formulas) and integrate term-by-term.

Answer

Use the identity \( \cos^2(x) = \frac{1+\cos(2x)}{2} \).

\[
\int \cos^2(x)\,dx=\int \frac{1+\cos(2x)}{2}\,dx
=\frac{1}{2}\int 1\,dx+\frac{1}{2}\int \cos(2x)\,dx
\]

\[
=\frac{x}{2}+\frac{1}{2}\cdot \frac{\sin(2x)}{2}+C
=\frac{x}{2}+\frac{\sin(2x)}{4}+C
\]

Final result: \( \frac{x}{2}+\frac{\sin(2x)}{4}+C \).

Detailed Explanation

We want to compute the indefinite integral:

\[
\int \cos^2(x)\,dx
\]

Step 1: Use a power-reduction identity.

To integrate \(\cos^2(x)\), it helps to rewrite it as a combination of \(\cos(2x)\) terms. A standard identity is:

\[
\cos^2(x)=\frac{1+\cos(2x)}{2}
\]

Step 2: Substitute the identity into the integral.

Replace \(\cos^2(x)\) with \(\frac{1+\cos(2x)}{2}\):

\[
\int \cos^2(x)\,dx=\int \frac{1+\cos(2x)}{2}\,dx
\]

Step 3: Pull out the constant factor.

Factor out \(\frac{1}{2}\):

\[
\int \frac{1+\cos(2x)}{2}\,dx=\frac{1}{2}\int \left(1+\cos(2x)\right)\,dx
\]

Step 4: Split the integral into two simpler integrals.

Use linearity of integrals:

\[
\frac{1}{2}\int \left(1+\cos(2x)\right)\,dx=\frac{1}{2}\left(\int 1\,dx+\int \cos(2x)\,dx\right)
\]

Step 5: Integrate each part.

First part:

\[
\int 1\,dx=x
\]

Second part: integrate \(\cos(2x)\). Let’s use the fact that:

\[
\int \cos(kx)\,dx=\frac{1}{k}\sin(kx)+C
\]

Here \(k=2\), so:

\[
\int \cos(2x)\,dx=\frac{1}{2}\sin(2x)
\]

Step 6: Combine everything and simplify.

Substitute the results back:

\[
\frac{1}{2}\left(\int 1\,dx+\int \cos(2x)\,dx\right)
=\frac{1}{2}\left(x+\frac{1}{2}\sin(2x)\right)+C
\]

Distribute \(\frac{1}{2}\):

\[
\frac{1}{2}\left(x+\frac{1}{2}\sin(2x)\right)
=\frac{x}{2}+\frac{1}{4}\sin(2x)
\]

So the final answer is:

\[
\int \cos^2(x)\,dx=\frac{x}{2}+\frac{1}{4}\sin(2x)+C
\]

See full solution

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Calculus FAQ

How do I integrate \( \cos^2(x) \) using a power-reduction identity?

Use \( \cos^2(x)=\frac{1+\cos(2x)}{2} \). Then \[ \int \cos^2(x)\,dx=\frac{1}{2}\int 1\,dx+\frac{1}{2}\int \cos(2x)\,dx=\frac{x}{2}+\frac{\sin(2x)}{4}+C. \]

What is a quick way using \( \sin(2x) \) form?

Since \( \sin(2x)=2\sin(x)\cos(x) \), from the identity above you get \[ \int \cos^2(x)\,dx=\frac{x}{2}+\frac{\sin(2x)}{4}+C. \]

Can I integrate \( \cos^2(x) \) by rewriting as \( 1-\sin^2(x) \)?

Use \( \cos^2(x)=1-\sin^2(x) \). Then \[ \int \cos^2(x)\,dx=\int 1\,dx-\int \sin^2(x)\,dx. \] With \( \int \sin^2(x)\,dx=\frac{x}{2}-\frac{\sin(2x)}{4}+C \), the result is \( \frac{x}{2}+\frac{\sin(2x)}{4}+C \).

What if the upper limit is used: compute \( \int_{0}^{\pi/2} \cos^2(x)\,dx \)?

Using \[ \int \cos^2(x)\,dx=\frac{x}{2}+\frac{\sin(2x)}{4}+C, \] evaluate: \[ \frac{\pi/2}{2}+\frac{\sin(\pi)}{4}-\left(0+\frac{\sin(0)}{4}\right)=\frac{\pi}{4}. \]

How do I check my answer by differentiating?

Differentiate \( \frac{x}{2}+\frac{\sin(2x)}{4} \): \[ \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{1}{2},\quad \frac{d}{dx}\left(\frac{\sin(2x)}{4}\right)=\frac{1}{4}\cdot 2\cos(2x)=\frac{\cos(2x)}{2}. \] Sum gives \( \frac{1+\cos(2x)}{2}=\cos^2(x). \)

What is the general method for \( \int \cos^n(x)\,dx \) when \( n \) is even?

For even powers, use identities like power-reduction for \( \cos^2(x) \). Often you rewrite \( \cos^n(x) \) into sums of cosines of multiples of \( x \) (via half-angle formulas) and integrate term-by-term.

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