Question Statement

Evaluate the definite integral:

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sin x\cos xdx$

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Background and Explanation

This problem involves integrating the product of sine and cosine functions over a specific interval. You can solve this using either the double-angle identity from trigonometry or by recognizing that $\cos x$ is the derivative of $\sin x$, allowing for a straightforward substitution.

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Solution

Method 1: Using the Double-Angle Identity

We begin by applying the trigonometric identity $\sin 2x=2\sin x\cos x$ to simplify the integrand.

$\begin{array}{rl}{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sin x\cos xdx& =\frac{1}{2}{\int }_{\pi /6}^{\pi /3}2\sin x\cos xdx\\ & =\frac{1}{2}{\int }_{\pi /6}^{\pi /3}\sin 2xdx\end{array}$

Now we integrate $\sin 2x$ using the standard integral formula $\int \sin axdx=-\frac{\cos ax}{a}$:

$\begin{array}{rl}& =\frac{1}{2}{\left[\frac{-\cos 2x}{2}\right]}_{\pi /6}^{\pi /3}\\ & =\frac{-1}{4}{\left[\cos 2x\right]}_{\pi /6}^{\pi /3}\end{array}$

Evaluate at the upper and lower limits:

$\begin{array}{rl}& =\frac{-1}{4}\left[\cos \left(2\cdot \frac{\pi }{3}\right)-\cos \left(2\cdot \frac{\pi }{6}\right)\right]\\ & =\frac{-1}{4}\left(\cos \frac{2\pi }{3}-\cos \frac{\pi }{3}\right)\end{array}$

Substitute the known values $\cos \frac{2\pi }{3}=-\frac{1}{2}$ and $\cos \frac{\pi }{3}=\frac{1}{2}$:

$\begin{array}{rl}& =\frac{-1}{4}\left(-\frac{1}{2}-\frac{1}{2}\right)\\ & =\frac{-1}{4}\left(\frac{-1-1}{2}\right)\\ & =\frac{-1}{4}\left(\frac{-2}{2}\right)\\ & =\frac{-1}{4}(-1)\\ & =\frac{1}{4}\end{array}$

Method 2: Using Substitution (Power Rule)

Alternatively, recognize that $\cos x$ is the derivative of $\sin x$. We can rewrite the integral as:

${\int }_{\pi /6}^{\pi /3}(\sin x)\cdot \cos xdx$

Using the formula $\int [f(x){]}^n{f}^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}$ where $f(x)=\sin x$, $f^{\prime }(x)=\cos x$, and $n=1$:

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

Substitute $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$ and $\sin \frac{\pi }{6}=\frac{1}{2}$:

$\begin{array}{rl}& =\frac{1}{2}\left[{\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2\right]\\ & =\frac{1}{2}\left(\frac{3}{4}-\frac{1}{4}\right)\\ & =\frac{1}{2}\left(\frac{3-1}{4}\right)\\ & =\frac{1}{2}\left(\frac{2}{4}\right)\\ & =\frac{1}{2}\cdot \frac{1}{2}\\ & =\frac{1}{4}\end{array}$

Both methods confirm that the value of the integral is $\boxed{\frac{1}{4}}$.

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Key Formulas or Methods Used

• Double-angle identity: $\sin 2x=2\sin x\cos x$
• Integration of sine: $\int \sin axdx=-\frac{\cos ax}{a}+C$
• Power rule for integration: $\int [f(x){]}^n{f}^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}+C$ (where $f^{\prime }(x)$ is the derivative of $f(x)$)
• Standard trigonometric values:
  • $\cos \frac{\pi }{3}=\frac{1}{2}$, $\cos \frac{2\pi }{3}=-\frac{1}{2}$
  • $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$, $\sin \frac{\pi }{6}=\frac{1}{2}$

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Summary of Steps

1. Identify the integral: Definite integral of $\sin x\cos x$ from $\frac{\pi }{6}$ to $\frac{\pi }{3}$
2. Method 1 - Double-angle: Multiply and divide by 2 to create $\sin 2x$, integrate to get $-\frac{\cos 2x}{2}$, evaluate at bounds
3. Method 2 - Substitution: Recognize $\cos xdx$ as $d(\sin x)$, apply power rule to get $\frac{{\sin }^{2}x}{2}$, evaluate at bounds
4. Calculate: Substitute trigonometric values and simplify to obtain final answer $\frac{1}{4}$