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