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Evaluate \[ \int_{0}^{\frac{\pi}{6}} \cos\left(3x - \frac{\pi}{6}\right) dx - Scottish Highers Maths - Question 11 - 2019
Question 11
Evaluate \[ \int_{0}^{\frac{\pi}{6}} \cos\left(3x - \frac{\pi}{6}\right) dx. \]
Worked Solution & Example Answer:Evaluate \[ \int_{0}^{\frac{\pi}{6}} \cos\left(3x - \frac{\pi}{6}\right) dx - Scottish Highers Maths - Question 11 - 2019
Step 1
Start to integrate
Answer
To integrate the function ( \cos(3x - \frac{\pi}{6}) ), we will use the substitution method. Let:
[ u = 3x - \frac{\pi}{6} ]
Then, the differential ( du = 3dx ) or ( dx = \frac{du}{3} ).
The limits change as follows:
When ( x = 0 ), ( u = 3(0) - \frac{\pi}{6} = -\frac{\pi}{6} )
When ( x = \frac{\pi}{6} ), ( u = 3 \times \frac{\pi}{6} - \frac{\pi}{6} = \frac{\pi}{3} ).
Therefore, the integral becomes:
[ \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \cos(u) \cdot \frac{du}{3} ]
Step 2
Complete integration
Answer
Now, we can integrate:
[ \int \cos(u) , du = \sin(u) + C ]
Thus, the integral evaluated from (-\frac{\pi}{6}) to (\frac{\pi}{3}) is:
[ \frac{1}{3}\left[ \sin(u) \right]_{-\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{1}{3}\left[ \sin\left(\frac{\pi}{3}\right) - \sin\left(-\frac{\pi}{6}\right) \right]
]
Step 3
Substitute limits
Answer
Now substituting the limits into the sine function:
[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} ]
Thus, we have:
[ \frac{1}{3}\left( \frac{\sqrt{3}}{2} - \left(-\frac{1}{2}\right) \right) = \frac{1}{3}\left( \frac{\sqrt{3}}{2} + \frac{1}{2} \right) = \frac{1}{3}\left( \frac{\sqrt{3} + 1}{2} \right) ]
Step 4