Question

\(∫^\frac{π}{2}_{\pi}{2}sin^7xdx\)

Answer 1

Let I=\(∫^\frac{π}{2}_{\pi}{2}sin^7xdx.....(1)\)

\(As sin^7(−x)=(sin(−x))^7=(−sinx)^7=−sin^7x,therefore,sin^2x is an odd function.\)

\(It is known that,if f(x)is an odd function,then ∫^a_-aƒ(x)dx=0\)

\(∴I=∫^\frac{π}{2}_\frac{π}{2}sin^7xdx=0\)