Question Statement

Q7. Evaluate the integral: $\int \frac{1+\cos 4t}{2}dt$

(Note: The original question contained $\cot 4t$, but the solution and recheck confirm the integrand is $\frac{1+\cos 4t}{2}$)

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Background and Explanation

This problem requires applying the linearity property of integrals and the standard integration formula for trigonometric functions of the form $\cos (ax)$. The expression $\frac{1+\cos 4t}{2}$ is related to the half-angle identity ${\cos }^2(2t)=\frac{1+\cos 4t}{2}$.

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Solution

Let $I$ denote the integral: $I=\int \frac{1+\cos 4t}{2}dt$

Factor out the constant $\frac{1}{2}$ from the integral: $\begin{array}{rl}I& =\frac{1}{2}\int (1+\cos 4t)dt\end{array}$

Apply the linearity property of integration, splitting the integral into two separate terms: $\begin{array}{rl}I& =\frac{1}{2}\int 1dt+\frac{1}{2}\int \cos 4tdt\end{array}$

Integrate each term using standard formulas. Recall that $\int \cos (ax)dx=\frac{\sin (ax)}{a}$:

• $\int 1dt=t$
• $\int \cos 4tdt=\frac{\sin 4t}{4}$

Substituting these results back: $\begin{array}{rl}I& =\frac{1}{2}t+\frac{1}{2}\cdot \frac{\sin 4t}{4}+c\end{array}$

Simplify the expression by multiplying the constants: $\begin{array}{rl}I& =\frac{t}{2}+\frac{\sin 4t}{8}+c\end{array}$

Verification by Differentiation

To confirm the result is correct, differentiate the solution with respect to $t$: $\begin{array}{rl}\frac{d}{dt}\left(\frac{t}{2}+\frac{\sin 4t}{8}+c\right)& =\frac{1}{2}\frac{d}{dt}(t)+\frac{1}{8}\frac{d}{dt}(\sin 4t)+\frac{d}{dt}(c)\end{array}$

Apply the chain rule to differentiate $\sin 4t$ (where the derivative of $\sin (u)$ is $\cos (u)\cdot u^{\prime }$, and $u=4t$, so $u^{\prime }=4$): $\begin{array}{rl}& =\frac{1}{2}\cdot 1+\frac{1}{8}\cdot \cos (4t)\cdot 4+0\\ & =\frac{1}{2}+\frac{4}{8}\cos 4t\\ & =\frac{1}{2}+\frac{1}{2}\cos 4t\\ & =\frac{1+\cos 4t}{2}\end{array}$

Since differentiation yields the original integrand, the solution $\frac{t}{2}+\frac{\sin 4t}{8}+c$ is verified as correct.

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Key Formulas or Methods Used

• Linearity of Integration: $\int [a\cdot f(t)+b\cdot g(t)]dt=a\int f(t)dt+b\int g(t)dt$
• Standard Integral: $\int \cos (ax)dx=\frac{\sin (ax)}{a}+c$
• Basic Integral: $\int 1dt=t+c$
• Chain Rule (for verification): $\frac{d}{dt}[\sin (at)]=a\cos (at)$
• Constant Multiple Rule: $\int k\cdot f(t)dt=k\int f(t)dt$

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Summary of Steps

1. Factor out the constant $\frac{1}{2}$ from the integrand
2. Split the integral into two parts: $\int 1dt$ and $\int \cos 4tdt$
3. Integrate the first term to get $t$
4. Integrate the second term using the formula for $\cos (ax)$ to get $\frac{\sin 4t}{4}$
5. Multiply both results by $\frac{1}{2}$ and combine
6. Simplify the fractions to obtain the final answer $\frac{t}{2}+\frac{\sin 4t}{8}+c$
7. Verify by differentiating the result to recover the original integrand $\frac{1+\cos 4t}{2}$