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प्रश्न
Evaluate: `int_0^(pi/2) sqrt(1 - cos 4x) "d"x`
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उत्तर
`int_0^(pi/2) sqrt(1 - cos 4x) "d"x`
= `int_0^(pi/2) sqrt(2sin^2 2x) "d"x` .......`[∵ 1 - cos theta = 2 sin^2 theta/2]`
= `sqrt(2) int_0^(pi/2) sin 2x "d"x`
= `sqrt(2)[(-cos 2x)/2]_0^(pi/2)`
= `sqrt(2)/2 [cos 2 pi/2 - cos 0]`
= `-(sqrt(2))/2 [cos pi - cos 0]`
= `-(sqrt(2))/2 (-1 - 1)`
= `sqrt(2)`
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