First slide

Evaluation of definite integrals

Question

$If I= \int_{- π / 2}^{π / 2} \sqrt{\cos x - \cos^{3} x} d x t h e n 3 I i s$

Moderate

A

1
B

2
C

3
D

4

Solution

Given integral
$\begin{matrix} I = \int_{- π / 2}^{π / 2} \sqrt{[\cos x (1 - \cos^{2} x)]} d x \\ = \int_{- π / 2}^{π / 2} \sqrt{(\cos x \sin^{2} x)} d x \end{matrix}$
$= \int_{- π / 2}^{π / 2} \sqrt{(\cos x)} | \sin x | d x - - - - - - - - (1)$
$Now, |\sin x| = \{\begin{matrix} - \sin x, if - π / 2 \leq x < 0 \\ \sin x, if 0 < x \leq π / 2 \end{matrix}$
$Thus, from equation (1), we have$
$I = \int_{- π / 2}^{0} \sqrt{(\cos x)} (- \sin x) d x + \int_{0}^{π / 2} \sqrt{(\cos x)} \sin x d x$
$Putting \cos x = t, - \sin x d x = d t, we get$
$\begin{matrix} I = \int_{0}^{1} t^{\frac{1}{2}} d t - \int_{1}^{0} t^{\frac{1}{2}} d t = 2 \int_{0}^{1} t^{\frac{1}{2}} d t \\ = 2 (\frac{2}{3}) {[t^{\frac{3}{2}}]}_{0}^{1} = \frac{4}{3} \end{matrix}$
$∴ 3 I = 4$

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