Question

The value of ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}dx$ is

• A. $0$
• B. $\frac{4}{3}$
• C. $\frac{2}{3}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B

$\frac{4}{3}$

Step 1: Compute the required value:

Given: ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}dx$

$$\begin{gathered} {\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sqrt{\cos \left(x\right)(1-{\cos }^2(x\left)\right)}dx \\ \Rightarrow {\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sqrt{\cos \left(x\right)\left({\sin }^2\right(x\left)\right)}dx\left[1-{\cos }^{2}x={\sin }^{2}x\right] \\ \Rightarrow {\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin \left(x\right)\sqrt{\cos \left(x\right)}dx \end{gathered}$$

Put $x=-x$

${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin (-x)\sqrt{\cos (-x)}dx=-{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin \left(x\right)\sqrt{\cos \left(x\right)}dx$

So ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}dx$ is an even function,

${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}dx=2{\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}dx$

Step 2: Integrate the expression

Let $\cos \left(x\right)=t$

$-\sin \left(x\right).dx=dt$

When $x=\frac{\mathrm{\pi }}{2},t=0$

$x=0,t=1$

$$\begin{gathered} 2{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}\right)dx=2{\int }_0^1\sqrt{t}dt \\ \Rightarrow 2{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}\right)dx=2{\int }_0^1{\left(t\right)}^{\frac{1}{2}}dt \\ \Rightarrow 2{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}\right)dx=2{\left|\frac{t^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}\right|}_0^1 \\ \Rightarrow 2{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}\right)dx=2{\left|\frac{t^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right|}_0^1 \\ \Rightarrow 2{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}\right)dx=\frac{4}{3}\left|1-0\right| \\ \Rightarrow 2{\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\sqrt{\cos \left(x\right)-{\cos }^3\left(x\right)}\right)dx=\frac{4}{3} \end{gathered}$$

Hence, option $\left(B\right)$ is the correct answer.