The integral ∫0π1+4sin2x2-4sinx2 dx equals - Infinity Learn by Sri Chaitanya

A
$4 \sqrt{3} - 4$
B
$4 \sqrt{3} - 4 - \frac{π}{3}$
C
$π - 4$
D
$\frac{2 π}{3} - 4 - 4 \sqrt{3}$

Solution:

Let $I = \int_{0}^{π} \sqrt{1 + 4 \sin^{2} \frac{x}{2} - 4 \sin \frac{x}{2}} d x$

$\begin{array}{l} I = \int_{0}^{π} \sqrt{{(1 - 2 \sin \frac{x}{2})}^{2}} d x \\ \Rightarrow I = \int_{0}^{π} |1 - 2 \sin \frac{x}{2}| d x \\ \Rightarrow I = \int_{0}^{π / 3} |1 - 2 \sin \frac{x}{2}| d x + \int_{π / 3}^{π} |1 - 2 \sin \frac{x}{2}| d x \\ \Rightarrow I = \int_{0}^{π / 3} (1 - 2 \sin \frac{x}{2}) d x + \int_{π / 3}^{π} (2 \sin \frac{x}{2} - 1) d x \\ \Rightarrow I = {[x + 4 \cos \frac{x}{2}]}_{0}^{π / 3} + {[- 4 \cos \frac{x}{2} - x]}_{π / 3}^{π} \\ \Rightarrow I = (\frac{π}{3} + 4 \cos \frac{π}{6} - 4) + (0 - π + 4 \cos \frac{π}{6} + \frac{π}{3} \\ \Rightarrow I = - \frac{π}{3} + 4 \sqrt{3} - 4 \end{array}$

The integral $\int_{0}^{π} \sqrt[]{1 + 4 \sin^{2} (\frac{x}{2}) - 4 \sin (\frac{x}{2})} d x$ equals

Solution:

Related content