Question

Evaluate the following integral:

\[\int_0^{2\pi} \sin^{100} x \cos^{101} xdx\]

 

Answer 1

\[\text{Let I }=\int_0^{2\pi} \sin^{100} x \cos^{101} xdx\]

Suppose

\[f\left( x \right) = \sin^{100} x \cos^{101} x\]

Now,

\[f\left( 2\pi - x \right) = \sin^{100} \left( 2\pi - x \right) \cos^{101} \left( 2\pi - x \right) = \left( - \sin x \right)^{100} \left( \cos x \right)^{101} = \sin^{100} x \cos^{101} x = f\left( x \right)\]

\[\therefore I = \int_0^{2\pi} \sin^{100} x \cos^{101} xdx = 2 \int_0^\pi \sin^{100} x \cos^{101} xdx ...................\left[ \int_0^{2a} f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( 2a - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( 2a - x \right) = - f\left( x \right)\end{cases} \right]\]

Again,

\[f\left( \pi - x \right) = \sin^{100} \left( \pi - x \right) \cos^{101} \left( \pi - x \right) = \left( \sin x \right)^{100} \left( - \cos x \right)^{101} = - \sin^{100} x \cos^{101} x = - f\left( x \right)\]

\[\therefore I = 2 \times 0 = 0 ...................\left[ \int_0^{2a} f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( 2a - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( 2a - x \right) = - f\left( x \right)\end{cases} \right]\]