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Find d∫0π41+sin2xdx

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प्रश्न

Find `int_0^(pi/4) sqrt(1 + sin 2x) "d"x`

योग
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उत्तर

We have I = `int_0^(pi/4) sqrt(1 + sin 2x) "d"x`

= `int_0^(pi/4) sqrt((sinx + cosx)^2) "d"x`

= `int_0^(pi/4) (sinx + cosx) "d"x`

= `(-cosx + sinx)_0^(pi/4)`

I = 1.

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अध्याय 7: Integrals - Solved Examples [पृष्ठ १५२]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 7 Integrals
Solved Examples | Q 12 | पृष्ठ १५२

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