Integrate the function \[\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\sqrt{\cos x-{{\cos }^{3}}x}dx=}\]

Hint: First take the integral assumed to be I. Now take the term \[\cos x\] common inside the square root. Use any trigonometric identity suitable which can simplify the square root of a product. Now divide the region of limits into 2 parts as negative and positive separately and then add them. This way you get the whole result from the given limits.

Complete step-by-step solution -
Given integral which we need to solve, is written as:
\[\Rightarrow \int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\sqrt{\cos x-{{\cos }^{3}}x}dx}\].
Assume this integral to be variable I, for every representation.
Now, take the term \[\cos x\] common inside the square root, we get:
\[\Rightarrow I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\sqrt{\cos x\left( 1-{{\cos }^{2}}x \right)}dx}\]
By using general algebraic identity, we can write it as:
\[\Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x=1\]
By substituting this equation into ‘I’, we can write it as:
\[\Rightarrow I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\sqrt{\cos x{{\sin }^{2}}x}dx}\]
By simplifying we can write the above equation in the form of:
\[\Rightarrow I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\sqrt{\cos x}.\sqrt{{{\sin }^{2}}x}dx}\]
By general mathematics, square root of a square can be said: \[\sqrt{{{x}^{2}}}=\left| x \right|\].
By substituting the above expression into I we can get it as:
\[\Rightarrow I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\sqrt{\cos x}.\left| \sin x \right|dx}\]
Now take limits 0 to \[\dfrac{\pi }{2}\] and \[\dfrac{-\pi }{2}\] to 0 separately. Integrals are P, Q.
Case 1: Limits are 0 to \[\dfrac{\pi }{2}\]. Let this integral be P.
\[\Rightarrow P=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\cos x}.\left| \sin x \right|dx}\]
In this region, the sine function is always positive. So, we can write the integral as:
\[\Rightarrow P=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\cos x}.\left| \sin x \right|dx}\]
Let us assume t = \[\cos x\], by differentiating this we get:
\[\Rightarrow dt=-\left( \sin x \right)dx\]. The limits will become as follows,
If x = 0, t = \[\cos x\] = 1; If \[x=\dfrac{\pi }{2}\], t = \[\cos x\] = 0.
So, integration after substituting them we get it as:
\[P=-\int\limits_{1}^{0}{\sqrt{t}dt}=\int\limits_{0}^{1}{\sqrt{t}dt}\] …..(1)
Case 2: Limits are \[\dfrac{-\pi }{2}\] to 0. Here sin is always negative. So, Q is:
\[\Rightarrow Q=-\int\limits_{\dfrac{-\pi }{2}}^{0}{\sqrt{\cos x}.\sin xdx}\]
Let us assume t = \[\cos x\], by differentiating this we get:
\[\Rightarrow dt=-\left( \sin x \right)dx\]. The limits will become as follows,
If x = 0, t = \[\cos x\] = 1;
 If \[x=\dfrac{-\pi }{2}\], t = \[\cos x\] = 0.
So, integration after substituting them we get it as:
\[\Rightarrow Q=\int\limits_{0}^{1}{\sqrt{t}dt}\] ……(2)
By adding the equation (1), (2) we get an equation as:
\[\Rightarrow P+Q=\int\limits_{0}^{1}{\sqrt{t}dt}+\int\limits_{0}^{1}{\sqrt{t}dt}\]
\[\Rightarrow P+Q=2\int\limits_{0}^{1}{\sqrt{t}dt}\]
By using P + Q = I, and \[\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+c\], we get the I as:
\[\Rightarrow I=\left[ 2\left( \dfrac{{{t}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right)+c \right]_{0}^{1}\]
By simplifying the equation given above, we can write it as:
\[\Rightarrow I=\left[ \dfrac{2{{t}^{\dfrac{3}{2}}}}{3}\times 2+c \right]_{0}^{1}\]
By substituting the limits we get the value of I as:
\[\Rightarrow I=\left[ \dfrac{4}{3}+c \right]-\left[ 0+c \right]=\dfrac{4}{3}\]
So the value of integration is \[\dfrac{4}{3}\].
Therefore option (b) is the correct answer to this question.

Note: Be careful while getting the relation between sin, \[\cos x\] as it is the base for the solution. An alternate is to use the given equation of sin as a quadratic equation. Find the value of \[\sin x\] by that you get \[\cos x\] value. You can use that directly. But it will be a very long process. First, you should try to diminish any trigonometric expression if not possible then use the brute force method. Some students forget the + 1 term in the equation, which will lead to wrong answers. So, don’t forget the term.