Question

Evaluate : ${\int }_0^{\pi }{\cos }^{3}xdx$

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is A

$0$

Evaluating the given integral :

Consider the given Equation as

$I={\int }_0^{\pi }{\cos }^{3}xdx......\left(1\right)$

When $x\to 0+\pi -x$

Substitute $-x$ for $x$ in the Equation $\left(1\right)$

$\begin{array}{rcl}I& =& {\int }_0^{\pi }{\cos }^3\left(-x\right)dx\\ I& =& -{\int }_0^{\pi }{\cos }^3\left(x\right)dx......\left(2\right)\end{array}$

Adding Equation $\left(1\right)\mathrm{and}\left(2\right)$ to get the value of $I$.

$\begin{array}{rcl}& \Rightarrow & 2I=\left({\int }_0^{\pi }{\cos }^{3}xdx\right)+\left(-{\int }_0^{\pi }{\cos }^{3}xdx\right)\\ & \Rightarrow & 2I={\int }_0^{\pi }{\cos }^{3}xdx-{\int }_0^{\pi }{\cos }^3\left(x\right)dx\\ & \Rightarrow & 2I=0\\ & \Rightarrow & I=\frac{0}{2}\\ & \Rightarrow & I=0\end{array}$

Therefore, the correct answer is Option (A)