Question Statement

Evaluate the definite integral:

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{1+\cos \theta }{(\theta +\sin \theta {)}^2}d\theta$

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Background and Explanation

This integral follows the pattern $\int [f(\theta ){]}^n\cdot f^{\prime }(\theta )d\theta$ where the numerator is the derivative of the denominator's base function. Recognize that $\frac{d}{d\theta }(\theta +\sin \theta )=1+\cos \theta$, allowing direct application of the power rule for integration without explicit substitution.

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Solution

We evaluate this integral by recognizing that the numerator $(1+\cos \theta )$ is exactly the derivative of the expression $(\theta +\sin \theta )$ appearing in the denominator.

First, rewrite the integrand to make the pattern clearer:

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}(\theta +\sin \theta {)}^{-2}\cdot (1+\cos \theta )d\theta$

Since $\frac{d}{d\theta }(\theta +\sin \theta )=1+\cos \theta$, we can apply the power rule for integration $\int [f(\theta ){]}^n\cdot f^{\prime }(\theta )d\theta =\frac{[f(\theta ){]}^{n+1}}{n+1}$ with $n=-2$:

$\begin{array}{rl}{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}(\theta +\sin \theta {)}^{-2}\cdot (1+\cos \theta )d\theta & ={\left[\frac{(\theta +\sin \theta {)}^{-2+1}}{-2+1}\right]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}\\ & ={\left[\frac{(\theta +\sin \theta {)}^{-1}}{-1}\right]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}\\ & =(-1){\left[\frac{1}{\theta +\sin \theta }\right]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}\end{array}$

Now evaluate at the upper and lower limits:

$\begin{array}{rl}& =(-1)\left[\frac{1}{\frac{\pi }{2}+\sin \frac{\pi }{2}}-\frac{1}{\frac{\pi }{6}+\sin \frac{\pi }{6}}\right]\\ & =(-1)\left[\frac{1}{\frac{\pi }{2}+1}-\frac{1}{\frac{\pi }{6}+\frac{1}{2}}\right]\\ & =(-1)\left[\frac{2}{\pi +2}-\frac{6}{\pi +3}\right]\\ & =\frac{-2}{\pi +2}+\frac{6}{\pi +3}\end{array}$

To obtain a numerical approximation (using $\pi \approx 3.14$):

$\begin{array}{rl}& =\frac{-2}{3.14+2}+\frac{6}{3.14+3}\\ & =\frac{-2}{5.14}+\frac{6}{6.14}\\ & \approx -0.3891+0.9771\\ & \approx 0.588\end{array}$

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Key Formulas or Methods Used

• Power Rule for Integration: $\int [f(x){]}^n\cdot f^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Special case: $\int \frac{f^{\prime }(x)}{[f(x){]}^2}dx=-\frac{1}{f(x)}+C$
• Fundamental Theorem of Calculus: ${\int }_a^b{f}^{\prime }(x)dx=f(b)-f(a)$

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Summary of Steps

1. Identify the pattern: Recognize that $1+\cos \theta =\frac{d}{d\theta }(\theta +\sin \theta )$
2. Rewrite the integrand: Express as $(\theta +\sin \theta {)}^{-2}(1+\cos \theta )$
3. Apply the power rule: Integrate to get $-\frac{1}{\theta +\sin \theta }$
4. Evaluate at bounds: Substitute $\theta =\frac{\pi }{2}$ and $\theta =\frac{\pi }{6}$
5. Simplify: Obtain $\frac{6}{\pi +3}-\frac{2}{\pi +2}$ (or combined as $\frac{4\pi +6}{(\pi +2)(\pi +3)}$)
6. Calculate numerically: Substitute $\pi \approx 3.14$ to get $\approx 0.588$