Question

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

Answer 1

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.