Question

The value of ${\int }_0^{\mathrm{\pi }}\left({\sin }^{50}\left(x\right)·{\cos }^{49}\left(x\right)\right)dx$ is

• A. $0$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{2}$
• D. $1$

Answer 1

The correct option is A

$0$

Explanation for the correct option

Given: ${\int }_0^{\mathrm{\pi }}\left({\sin }^{50}\left(x\right)·{\cos }^{49}\left(x\right)\right)dx$

Let $I={\int }_0^{\mathrm{\pi }}\left({\sin }^{50}\left(x\right)·{\cos }^{49}\left(x\right)\right)dx------\left(1\right)$

using the property $x\to a+b-x$ ${\int }_b^{a}x·dx={\int }_b^a(a+b-x)·dx$

$$\begin{gathered} I={\int }_0^{\mathrm{\pi }}\left({\sin }^{50}(\mathrm{\pi }+0-x)·{\cos }^{49}(\mathrm{\pi }+0-\mathrm{x})\right)dx \\ \Rightarrow I=-{\int }_0^{\mathrm{\pi }}\left({\sin }^{50}\left(x\right)·{\cos }^{49}\left(x\right)\right)dx------\left(2\right) \end{gathered}$$

Adding equations $\left(1\right)$ and $\left(2\right)$

$$\begin{gathered} \Rightarrow I+I={\int }_0^{\mathrm{\pi }}\left({\sin }^{50}\left(x\right)·{\cos }^{49}\left(x\right)\right)dx-{\int }_0^{\mathrm{\pi }}\left({\sin }^{50}\left(x\right)·{\cos }^{49}\left(x\right)\right)dx \\ \Rightarrow 2I={\int }_0^{\mathrm{\pi }}{\sin }^{50}\left(x\right)·{\cos }^{49}\left(x\right)-\left({\sin }^{50}\left(x\right)·{\cos }^{49}\left(x\right)\right)dx \\ \Rightarrow 2I=0 \\ \Rightarrow I=0 \end{gathered}$$

Hence, option A is correct.