Resources for Trig Integration
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Questions
16
With Worked SolutionClick Here -
Video Tutorials
3
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HSC Questions
7
With Worked SolutionClick Here
Trig Integration Theory
![\textbf{Trigonometric Functions}\\ When integrating trigonometry functions it is important to recall identities such as:\\ $\begin{aligned} & \cos 2 \theta=2 \cos ^2 \theta-1 \\ & \cos 2 \theta=1-2 \sin ^2 \theta \\ & \sin \theta \cos \varphi=\dfrac{1}{2}(\sin (\theta+\varphi)+\sin (\theta-\varphi)) \\ & \cos \theta \cos \varphi=\dfrac{1}{2}(\cos (\theta+\varphi)+\cos (\theta-\varphi)) \\ & \cos \theta \sin \varphi=\dfrac{1}{2}(\sin (\theta+\varphi)-\sin (\theta-\varphi)) \\ & \sin \theta \sin \varphi=\dfrac{1}{2}(\cos (\theta-\varphi)-\cos (\theta+\varphi)) \end{aligned}$\\ \begin{multicols}{2} \textbf{Example 1}\\ %Question 12023 Evaluate \(\displaystyle \int\limits_0^{\frac{\pi }{6}} {{{\cos }^2}\theta \,d\theta } \) \\ \textbf{Example 1 solution}\\ $\begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{6}} \cos ^{2} \theta \, d \theta &=\displaystyle \int_{0}^{\frac{\pi}{6}} \frac{1}{2}(\cos 2\theta+1) \, d \theta \\ &=\frac{1}{2}\left[\frac{1}{2} \sin 2 \theta+\theta\right]_{0}^{\frac{\pi}{6}} \\ &=\frac{1}{2}\left[\frac{1}{2} \sin \frac{\pi}{3}+\frac{\pi}{6}-0 \right]\\ &=\frac{1}{4} \times \frac{\sqrt{3}}{2}+\frac{\pi}{12} \\ &=\frac{\sqrt{3}}{8}+\frac{\pi}{12} \end{aligned}$\\ \columnbreak \textbf{Example 2}\\ %Question 12024 Evaluate \(\displaystyle \int\limits_0^{\frac{\pi }{3}} {\cos 3x\,\sin x\,dx} \)\\ \textbf{Example 2 solution}\\ $\begin{aligned} &\displaystyle \int_{0}^{\frac{\pi}{3}} \cos 3 x \sin x \,dx=\frac{1}{2} \displaystyle \int_{0}^{\frac{\pi}{3}} \sin (3 x+x)-\sin (3 x-x) \,dx\\ \end{aligned}$\\ $\begin{aligned} I&=\frac{1}{2} \displaystyle \int_{0}^{\frac{\pi}{3}} \sin 4 x-\sin 2 x \,dx \\ &=\frac{1}{2}\left[-\frac{1}{4} \cos 4 x+\frac{1}{2} \cos 2 x\right]_{0}^{\frac{\pi}{3}} \\ &=\frac{1}{2}\left[-\frac{1}{4} \cos \frac{4 \pi}{3}+\frac{1}{2} \cos \frac{2 \pi}{3}-\left(-\frac{1}{4} \cos 0+\frac{1}{2} \cos 0\right)\right] \\ &=\frac{1}{2}\left[-\frac{1}{4} \times-\frac{1}{2}+\frac{1}{2} \times-\frac{1}{2}+\frac{1}{4}-\frac{1}{2}\right] \\ &=\frac{1}{2}\left[\frac{1}{8}-\frac{1}{4}+\frac{1}{4}-\frac{1}{2}\right] \\ &=-\frac{3}{16} \end{aligned}$\\](/media/2pnb1z3q/trig-integration.png)
