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Integrate the function: cosx4-sin2x

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प्रश्न

Integrate the function:

`cos x/sqrt(4 - sin^2 x)`

योग
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उत्तर

Let I = `int (cos x)/sqrt(4 - sin^2 x)`dx

Substituting sin x = 2t

cos x dx = 2 dt

Hence,  `I = int (2 dt)/sqrt(4 - 4t^2)`

`= int (2 dt)/(2sqrt(1 - t^2))`

`= int 1/sqrt(1 - t^2) dt = sin^-1 t + C`

`= sin^-1 ((sin x)/2) + C      ....[(because sin x = 2 t), (=> t = (sin x)/2)]`

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अध्याय 7: Integrals - Exercise 7.12 [पृष्ठ ३५२]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.12 | Q 9 | पृष्ठ ३५२

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