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Calculus, 2ADV C4 2023 MET1 5

  1. Evaluate  \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\).   (1 mark)
  2. Hence, or otherwise, find all values of \(k\) such that \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx=\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\), where \(-3\pi<k<2\pi\).   (3 marks)
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a.    \(\dfrac{1}{2}\)

b.    \(k=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Show Worked Solution
a.    \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\) \(=\left[-\cos x\right]_0^\frac{\pi}{3}\)
    \(=-\cos\dfrac{\pi}{3}+\cos 0\)
    \(=-\dfrac{1}{2}+1\)
    \(=\dfrac{1}{2}\)

 

b.    \(\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\) \(=\left[\sin x\right]_k^\frac{\pi}{2}\)
    \(=\sin\bigg(\dfrac{\pi}{2}\bigg)-\sin (k)\)
    \(=1-\sin (k)\)

 
\(\text{Using part (a):}\)

\(1-\sin (k)\) \(=\dfrac{1}{2}\)
\(\sin (k)\) \(=\dfrac{1}{2}\)
\(\therefore\ k\) \(=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Calculus, 2ADV C4 2014 HSC 13a

  1.  Differentiate  `3 + sin 2x`.    (1 mark)

  2.  Hence, or otherwise, find  `int (cos2x)/(3 + sin 2x)\ dx`.    (2 marks)

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  1. `2 cos 2x`
  2. `ln (3 + sin 2x)^(1/2) + C`
Show Worked Solution
i. `y` `= 3 + sin 2x`
  `dy/dx` `= 2 cos 2x`

 

ii. `int (cos 2x)/(3 + sin 2x)\ dx`
  `= 1/2 int (2 cos 2x)/(3 + sin 2x)\ dx`
  `= 1/2  ln (3 + sin 2x) + C\ \ \ \ \ text{(from part (i))}`