Question

Evaluate the value of $\int\limits_0^\pi {{{\cos }^3}xdx} $.
A. 0
B. 1
C. -1
D. $\dfrac{1}{{2\sqrt 2 }}$

Answer 1

Hint: We will apply the definite integral property to rewrite the given definite integral and simplify it by using trigonometry ratios of supplementary angle. Add both of them to get the value of the given integral.

Formula Used:
$\int\limits_b^a {f\left( x \right)dx} = \int\limits_b^a {f\left( {a - x} \right)dx} $

Complete step by step solution:
Given integration $\int\limits_0^\pi {{{\cos }^3}xdx} $.
Let $I = \int\limits_0^\pi {{{\cos }^3}xdx} $ ……(i)
Apply the formula $\int\limits_b^a {f\left( x \right)dx} = \int\limits_b^a {f\left( {a - x} \right)dx} $
$I = \int\limits_0^\pi {{{\cos }^3}\left( {\pi - x} \right)dx} $
We know $\cos \left( {\pi - x} \right) = - \cos x$.
$I = - \int\limits_0^\pi {\cos xdx} $ ….(ii)
Add equation (i) and (ii)
$I + I = \int\limits_0^\pi {{{\cos }^3}xdx} - \int\limits_0^\pi {{{\cos }^3}xdx} $
$ \Rightarrow 2I = 0$
$ \Rightarrow I = 0$

Option ‘A’ is correct

Additional information:
If an integral has an upper limit and a lower limit, then the integration is known as a definite integral.
some properties of the definite integral:
$\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( t \right)dt} $
$\int_a^b {f\left( x \right)dx} = \int_b^a {f\left( x \right)dx} $
$\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( {a + b - x} \right)dx} $
$\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} $

Note: To solve the question, you need to know all properties of the definite integral. Since the lower limit of the given integration is zero, so we will apply $\int\limits_b^a {f\left( x \right)dx} = \int\limits_b^a {f\left( {a - x} \right)dx} $. Then we will use the supplementary angle to simplify it and add them.