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D∫sinx3+4cos2xdx = ______. - Mathematics

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प्रश्न

`int sinx/(3 + 4cos^2x) "d"x` = ______.

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उत्तर

`int sinx/(3 + 4cos^2x) "d"x` = `- 1/(2sqrt(3)) tan^-1 (2/sqrt(3) cos x) + "C"`.

Explanation:

Let I = `int sinx/(3 + 4cos^2x) "d"x` 

Put cos x = t

∴ – sin x dx = dt

⇒ sinx dx = – dt

∴ I = `- int "dt"/(3 + 4"t"^2)`

= `- 1/4 int "dt"/(3/4 + "t"^2)`

= `- 1/4 int "dt"/((sqrt(3)/2)^2 + "t"^2)`

= `1/4 xx 1/(sqrt(3)/2) tan^-1 ("t"/(sqrt(3)/2)) + "C"`

= ` 1/(2sqrt(3)) tan^-1 ((2"t")/sqrt(3)) + "C"`

= `- 1/(2sqrt(3)) tan^-1 ((2cosx)/sqrt(3)) + "C"`

Hence I = `- 1/(2sqrt(3)) tan^-1 (2/sqrt(3) cos x) + "C"`.

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अध्याय 7: Integrals - Exercise [पृष्ठ १६९]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 7 Integrals
Exercise | Q 62 | पृष्ठ १६९

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