Question

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

Answer 1

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

व्याख्या:

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

cos x = t रखिए

∴ – 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"`

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