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This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.

$$\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}}\text{ d}t\text{.}$$ Because of the $\sqrt{t^2-1} = \sqrt{t^2-1^2}$ in the integrand, I set $t = 1\sec(\theta) = \sec(\theta)$, and $\theta = \sec^{-1}(t)$. Thus $\text{d}t = \sec(\theta)\tan(\theta)\text{ d}\theta$ and $$\begin{align}\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}} \text{ d}t&= \int\limits_{\pi/4}^{\pi/3}\dfrac{1}{\sec^{3}(\theta)\sqrt{\sec^{2}(\theta)-1}}\sec(\theta)\tan(\theta)\text{ d}\theta \\ &= \int\limits_{\pi/4}^{\pi/3}\dfrac{1}{\sec^{2}(\theta)}\text{ d}\theta \\ &= \int\limits_{\pi/4}^{\pi/3}\cos^{2}(\theta)\text{ d}\theta \\ &= \dfrac{1}{2}\int\limits_{\pi/4}^{\pi/3}[1+\cos(2\theta)]\text{ d}\theta \\ &= \dfrac{1}{2}\left\{\dfrac{\pi}{3}-\dfrac{\pi}{4}+\dfrac{1}{2}\left[\sin\left(\dfrac{2\pi}{3}\right) - \sin\left(\dfrac{\pi}{2}\right)\right]\right\}\\ &= \dfrac{1}{2}\left[\dfrac{\pi}{12}+\dfrac{1}{2}\left(\dfrac{\sqrt{3}}{2}-1\right)\right] \\ &= \dfrac{\pi}{24}+\dfrac{1}{4}\left(\dfrac{\sqrt{3}-2}{2}\right)\text{.} \end{align}$$

A quick side question: without access to a calculator and assuming I have the values of $\sin(x)$ memorized for $x = 0, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}$, what is the easiest way to find $\sin\left(\dfrac{2\pi}{3}\right)$?

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  • $\begingroup$ $sin(x) = cos(x - \pi/2)$, then $sin(2\pi/3) = cos(\pi/6)$ $\endgroup$
    – user67133
    Oct 19, 2014 at 1:38
  • $\begingroup$ @user01123581321345589144... - and there's another identity to add to my formula sheet. Thank you! $\endgroup$ Oct 19, 2014 at 1:39
  • $\begingroup$ The (equivalent) identity $\sin(x)=\cos(\pi/2-x)$ (and since $\cos$ is even, we can get the previous one from there) might be easier to remember, since $\pi/2-x$ is the complementary angle to $x$ and "cosine" is the sine of complement. $\endgroup$ Oct 19, 2014 at 1:42
  • $\begingroup$ @Meelo - Clever! :O Thank you! $\endgroup$ Oct 19, 2014 at 1:43

2 Answers 2

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Your working looks alright.$$\sin \theta= \sin (\pi - \theta)$$ Is another identity.

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There are no issues with your solution. You can easily find $\sin \frac{2\pi}{3}$ using $\sin x = \sin (\pi - x)$ for $x = \frac{\pi}{3}$.

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