Question Statement

Evaluate the integral:

$\int \frac{{\cos }^{2}z}{7}dz$

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Background and Explanation

This problem requires integrating a trigonometric function raised to a power. Since ${\cos }^{2}z$ cannot be integrated directly using basic rules, we use a power-reduction identity (double-angle formula) to rewrite ${\cos }^{2}z$ in terms of $\cos 2z$, which is readily integrable.

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Solution

Let $I=\int \frac{{\cos }^{2}z}{7}dz$.

Step 1: Factor out the constant

Using the constant multiple rule for integration, we factor out $\frac{1}{7}$:

$I=\frac{1}{7}\int {\cos }^{2}zdz$

Step 2: Apply the double-angle identity

We use the identity ${\cos }^{2}x=\frac{1+\cos 2x}{2}$ to convert the squared cosine into a linear term:

$I=\frac{1}{7}\int \left(\frac{1+\cos 2z}{2}\right)dz$

Step 3: Simplify and split the integral

Multiply the constants and separate the integral into two simpler integrals:

$I=\frac{1}{14}\int (1+\cos 2z)dz=\frac{1}{14}\int 1dz+\frac{1}{14}\int \cos 2zdz$

Step 4: Integrate term by term

• The integral of $1$ with respect to $z$ is $z$
• For $\int \cos 2zdz$, we use the rule $\int \cos (ax)dx=\frac{\sin (ax)}{a}$ with $a=2$:

$I=\frac{1}{14}z+\frac{1}{14}\left(\frac{\sin 2z}{2}\right)+c$

Step 5: Final simplification

Combine the fractions to get the final result:

$I=\frac{z}{14}+\frac{\sin 2z}{28}+c$

Verification (Recheck):

To confirm our answer is correct, we differentiate the result with respect to $z$:

$\frac{d}{dz}\left(\frac{z}{14}+\frac{\sin 2z}{28}+c\right)=\frac{1}{14}+\frac{1}{28}\cdot (2\cos 2z)+0$

$=\frac{1}{14}+\frac{2\cos 2z}{28}=\frac{1}{14}+\frac{\cos 2z}{14}=\frac{1+\cos 2z}{14}$

Factoring out $\frac{1}{7}$ and applying the identity in reverse:

$=\frac{1}{7}\left(\frac{1+\cos 2z}{2}\right)=\frac{1}{7}{\cos }^{2}z$

This matches the original integrand, confirming our solution is correct.

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Key Formulas or Methods Used

• Constant Multiple Rule: $\int k\cdot f(x)dx=k\int f(x)dx$
• Power-Reduction Identity: ${\cos }^{2}x=\frac{1+\cos 2x}{2}$
• Integration of Cosine: $\int \cos (ax)dx=\frac{\sin (ax)}{a}+c$
• Basic Integration: $\int 1dx=x+c$
• Linearity of Integration: $\int [f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$

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Summary of Steps

1. Factor out the constant $\frac{1}{7}$ from the integral
2. Apply the double-angle identity ${\cos }^{2}z=\frac{1+\cos 2z}{2}$ to eliminate the square
3. Distribute the $\frac{1}{14}$ and split into two separate integrals
4. Integrate each term: $\int 1dz=z$ and $\int \cos 2zdz=\frac{\sin 2z}{2}$
5. Combine results and add the constant of integration $c$
6. (Optional) Verify by differentiating the result to recover the original integrand