Aussie Maths & Science Teachers: Save your time with SmarterEd
Evaluate \(\displaystyle\int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} \cos ^2(3 x) d x\). (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
\(\dfrac{\pi}{12}\)
| \(\displaystyle\int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} \cos ^2(3x) dx\) | \(=\displaystyle \dfrac{1}{2} \int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} (\cos6x+1) dx \) |
| \(=\dfrac{1}{2}\left[\dfrac{1}{6} \sin 6 x+x\right]_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}}\) | |
| \(=\dfrac{1}{2}\left[\left(\dfrac{1}{6} \sin 2 \pi+\dfrac{\pi}{3}\right)-\left(\dfrac{1}{6} \sin \pi+\dfrac{\pi}{6}\right)\right]\) | |
| \(=\dfrac{1}{2}\left(\dfrac{\pi}{3}-\dfrac{\pi}{6}\right)\) | |
| \(=\dfrac{\pi}{12}\) |
--- 6 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) ---
Show Answers Only
i.
\(\text{LHS}\)
\(=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2\)
\(=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)\)
\(=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)\)
\(=\dfrac{1+\cos ^2(2 x)}{2}\)
ii. \(\dfrac{3 \pi}{16}\)Show Worked Solution
i.
\(\text{LHS}\)
\(=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2\)
\(=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)\)
\(=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)\)
\(=\dfrac{1+\cos ^2(2 x)}{2}\)
ii.
\(\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x\)
\(=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\cos ^2(2 x) d x\)
\(=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\dfrac{1}{2}(1+\cos (4 x)) d x\)
\(=\dfrac{1}{2}\left[\dfrac{3}{2}x +\dfrac{1}{8} \sin (4 x)\right]_0^{\frac{\pi}{4}}\)
\(=\dfrac{1}{2}\left[\dfrac{3}{2} \times \dfrac{\pi}{4}+\dfrac{1}{8} \sin \pi-0\right]\)
\(=\dfrac{3 \pi}{16}\)