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Evaluate: ∫0π21-cos4x dx - Mathematics and Statistics

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प्रश्न

Evaluate: `int_0^(pi/2) sqrt(1 - cos 4x)  "d"x`

योग
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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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Methods of Evaluation and Properties of Definite Integral
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2.4: Definite Integration - Short Answers I

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