Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
$\int\limits^{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
✓
Answer
We have,
$\int^\limits{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\frac{\sin2\theta}{\cos^32\theta}\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta$
Let $\tan2\theta=\text{t}$
Differentiating w.r.t. x, we get
$2\sec^22\theta\text{d}\theta=\text{dt}$
Now, $\theta=0\Rightarrow\text{t}=0$
$\theta=\frac{\pi}{6}\Rightarrow\text{t}=\sqrt{3}$
$\therefore\ \int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta=\frac{1}{2}\int^\limits{\sqrt{3}}_0\text{t dt}=\frac{1}{2}\Big[\frac{\text{t}^2}{2}\Big]^{\sqrt{\text{3}}}_0$
$=\frac{3}{4}$
$\int^\limits{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\frac{\sin2\theta}{\cos^32\theta}\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta$
Let $\tan2\theta=\text{t}$
Differentiating w.r.t. x, we get
$2\sec^22\theta\text{d}\theta=\text{dt}$
Now, $\theta=0\Rightarrow\text{t}=0$
$\theta=\frac{\pi}{6}\Rightarrow\text{t}=\sqrt{3}$
$\therefore\ \int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta=\frac{1}{2}\int^\limits{\sqrt{3}}_0\text{t dt}=\frac{1}{2}\Big[\frac{\text{t}^2}{2}\Big]^{\sqrt{\text{3}}}_0$
$=\frac{3}{4}$
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