Q109 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\int_\limits{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sqrt{\cos x-\cos ^3 x} d x \text { is equal to }$
- A0
- B$-\dfrac{2}{3}$
- C$-\dfrac{1}{3}$
- Correct answerD$\dfrac{4}{3}$
Solution
The integral is $I = \int_{-\pi/2}^{\pi/2} \sqrt{\cos x - \cos^3 x} dx$.
Since $\cos x - \cos^3 x = \cos x(1 - \cos^2 x) = \cos x \sin^2 x$, the integrand becomes $\sqrt{\cos x \sin^2 x} = |\sin x| \sqrt{\cos x}$.
The integral is $I = \int_{-\pi/2}^{\pi/2} |\sin x| \sqrt{\cos x} dx$.
Since the integrand $f(x) = |\sin x| \sqrt{\cos x}$ is an even function because $f(-x) = |\sin(-x)| \sqrt{\cos(-x)} = |-\sin x| \sqrt{\cos x} = |\sin x| \sqrt{\cos x} = f(x)$, we can write $I = 2 \int_{0}^{\pi/2} \sin x \sqrt{\cos x} dx$.
Let $u = \cos x$, then $du = -\sin x dx$. When $x = 0$, $u = 1$. When $x = \pi/2$, $u = 0$.
$I = 2 \int_{1}^{0} \sqrt{u} (-du) = 2 \int_{0}^{1} u^{1/2} du$.
$I = 2 \left[ \dfrac{u^{3/2}}{3/2} \right]_{0}^{1} = 2 \times \dfrac{2}{3} [u^{3/2}]_{0}^{1} = \dfrac{4}{3} (1 - 0) = \dfrac{4}{3}$.
Answer: $\dfrac{4}{3}$
Since $\cos x - \cos^3 x = \cos x(1 - \cos^2 x) = \cos x \sin^2 x$, the integrand becomes $\sqrt{\cos x \sin^2 x} = |\sin x| \sqrt{\cos x}$.
The integral is $I = \int_{-\pi/2}^{\pi/2} |\sin x| \sqrt{\cos x} dx$.
Since the integrand $f(x) = |\sin x| \sqrt{\cos x}$ is an even function because $f(-x) = |\sin(-x)| \sqrt{\cos(-x)} = |-\sin x| \sqrt{\cos x} = |\sin x| \sqrt{\cos x} = f(x)$, we can write $I = 2 \int_{0}^{\pi/2} \sin x \sqrt{\cos x} dx$.
Let $u = \cos x$, then $du = -\sin x dx$. When $x = 0$, $u = 1$. When $x = \pi/2$, $u = 0$.
$I = 2 \int_{1}^{0} \sqrt{u} (-du) = 2 \int_{0}^{1} u^{1/2} du$.
$I = 2 \left[ \dfrac{u^{3/2}}{3/2} \right]_{0}^{1} = 2 \times \dfrac{2}{3} [u^{3/2}]_{0}^{1} = \dfrac{4}{3} (1 - 0) = \dfrac{4}{3}$.
Answer: $\dfrac{4}{3}$
Hence, the correct answer is option: 'D'.