Question

The value of ${\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\sin }^{100}\left(x\right)-{\cos }^{100}\left(x\right)\right)dx$ is

• A. $\frac{1}{100}$
• B. $\frac{100!}{{\left(100\right)}^{100}}$
• C. $\frac{\mathrm{\pi }}{100}$
• D. $0$

Answer 1

The correct option is D

$0$

Explanation for correct option

Given: ${\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\sin }^{100}\left(x\right)-{\cos }^{100}\left(x\right)\right)dx$

Let $I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\sin }^{100}\left(x\right)-{\cos }^{100}\left(x\right)\right)dx$

$I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\sin }^{100}\left(x\right)-{\cos }^{100}\left(x\right)\right)dx-----\left(1\right)$

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

$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\sin }^{100}\left(\frac{\mathrm{\pi }}{2}+0-x\right)-{\cos }^{100}\left(\frac{\mathrm{\pi }}{2}+0-x\right)\right)dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\sin }^{100}\left(\frac{\mathrm{\pi }}{2}-x\right)-{\cos }^{100}\left(\frac{\mathrm{\pi }}{2}-x\right)\right)dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\cos }^{100}\left(x\right)-{\sin }^{100}\left(x\right)\right)dx-----\left(2\right) \end{gathered}$$

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

$$\begin{gathered} \Rightarrow I+I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\sin }^{100}\left(x\right)-{\cos }^{100}\left(x\right)\right)dx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\cos }^{100}\left(x\right)-{\sin }^{100}\left(x\right)\right)dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left({\sin }^{100}\left(x\right)-{\cos }^{100}\left(x\right)-{\sin }^{100}\left(x\right)+{\cos }^{100}\left(x\right)\right)·dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}0·dx \\ \Rightarrow \mathit{2}I\mathit{=}\mathit{0} \\ \Rightarrow I\mathit{=}\mathit{0} \end{gathered}$$

Hence, option D is correct.