By using the properties of definite integrals, evaluate the integral ∫π2−π2sin7xdx
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Solution
Let I=∫π2−π2sin7xdx .......... (1) As sin7(−x)=(sin(−x))7=(−sinx)7=−sin7x, Therefore, sin7x is an odd function. It is known that, if f(x) is an odd function, then ∫a−af(x)dx=0 ∴I=∫π2−π2sin7xdx=0