Don't Commit Careless Mistake

Calculus Level 4

$\displaystyle \int_{0}^{\pi} \sqrt{1+4 \sin^2\left( \frac{x}{2} \right) -4 \sin\left( \frac{x}{2} \right)} \ \mathrm{d}x = \ ?$

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\pi-4

\frac{2\pi}{3}-4-4 \sqrt{3}

4 \sqrt{3}-4

4 \sqrt{3}-4-\frac{\pi}{3}

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Brian Charlesworth
Mar 9, 2015

The expression under the root sign is $(2\sin(\frac{x}{2}) - 1)^{2}$ , so the integral becomes

$\displaystyle\int_{0}^{\pi} |(2\sin(\frac{x}{2}) - 1)| dx.$

Now $\sin(\frac{x}{2}) \le \frac{1}{2}$ on the interval $[0, \frac{\pi}{3}]$ and $\sin(\frac{x}{2}) \ge \frac{1}{2}$ on the interval $[\frac{\pi}{3}, \pi]$ , so we need to evaluate the integral in two parts, i.e.,

$\displaystyle\int_{0}^{\frac{\pi}{3}} (1 - 2\sin(\frac{x}{2})) dx + \int_{\frac{\pi}{3}}^{\pi} (2\sin(\frac{x}{2}) - 1) dx =$

$(x + 4\cos(\frac{x}{2}))|_{x=0}^{\frac{\pi}{3}} - (x + 4\cos(\frac{x}{2}))|_{x=\frac{\pi}{3}}^{\pi} =$

$(\frac{\pi}{3} + 4\cos(\frac{\pi}{6}) - 4) - (\pi - \frac{\pi}{3} - 4\cos(\frac{\pi}{6})) = \boxed{4\sqrt{3} - 4 - \frac{\pi}{3}}.$

exact. The careless mistake here is assuming $\sqrt{x^2}=x$ . On the basis of this wrong assumption, the answer will come out to be different, which is also given in the options.

But the correct one is $\sqrt{x^2}=|x|$ , leading to the correct, as solved by @brian charlesworth sir.

Sandeep Bhardwaj - 6 years, 3 months ago

Don't Commit Careless Mistake

Practice the set Target JEE_Advanced - 2015 and boost up your preparation.

Image Credit: Wikimedia Reflection angles

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