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int0^(2pi)sin^(100)xcos^(99)x dx...

`int_0^(2pi)sin^(100)xcos^(99)x dx`

Text Solution

Verified by Experts

The correct Answer is:
`0`
`I=int_(0)^(2pi)sin^(100)xcos^(99)x dx`
Here `f(x)=sin^(100)x cos^(99) x` for which `f(2pi-x)=f(x)`
`:.I=2int_(0)^(pi)sin^(100)xcos^(99)xdx`
`=2int_(0)^(pi)sin^(100)(pi-x)cos^(99)(pi-x)dx`
`=-2int_(0)^(pi)sin^(100)xcos^(99)xdx`
`=-I`
or `2I=` or `I=0`
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