Question

${\int }_0^{2\pi }{\sin }^6\left(x\right){\cos }^5\left(x\right)dx=$

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

Answer 1

The correct option is C

$0$

Explanation for the correct option:

Integrating by changing the variables:

Let $I={\int }_0^{2\pi }{\sin }^6\left(x\right){\cos }^5\left(x\right)dx$

$$\begin{gathered} ={\int }_0^{2\pi }{\sin }^6\left(x\right){\left({\cos }^2\left(x\right)\right)}^2\cos \left(x\right)dx \\ ={\int }_0^{2\pi }{\sin }^6\left(x\right){\left(1-{\sin }^2\left(x\right)\right)}^2\cos \left(x\right)dx \end{gathered}$$

Substituting $\sin \left(x\right)=t$,

Therefore, by differentiating on both sides, we get

$\cos \left(x\right)dx=dt$

At $x=0,\sin \left(0\right)=0$

At $x=2\pi ,\sin \left(2\pi \right)=0$

By changing the limits we get,

$$\begin{gathered} I={\int }_0^0{t}^6{\left(1-t^2\right)}^{2}dt \\ I=0\left[\because {\int }_a^{a}f\left(x\right)dx=0\right] \end{gathered}$$

Therefore, the value of ${\int }_0^{2\pi }{\sin }^6\left(x\right){\cos }^5\left(x\right)dx=0$

Hence, option C is the correct answer.