Question:
\(∫^\frac{π}{2}_{\pi}{2}sin^7xdx\)
\(∫^\frac{π}{2}_{\pi}{2}sin^7xdx\)
Updated On: Sep 17, 2023
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Solution and Explanation
Let I=\(∫^\frac{π}{2}_{\pi}{2}sin^7xdx.....(1)\)
\(As sin^7(−x)=(sin(−x))^7=(−sinx)^7=−sin^7x,therefore,sin^2x is an odd function.\)
\(It is known that,if f(x)is an odd function,then ∫^a_-aƒ(x)dx=0\)
\(∴I=∫^\frac{π}{2}_\frac{π}{2}sin^7xdx=0\)
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