Advertisements
Advertisements
प्रश्न
Find `int "dx"/(2sin^2x + 5cos^2x)`
उत्तर
Dividing numerator and denominator by cos2x, we have
I = `int (sec^2x "d"x)/(2tan^2x + 5)`
Put tanx = t
So that sec2x dx = dt.
Then I = `int "dt"/(2"t"^2 + 5) = 1/2 int "dt"/("t"^2 + (sqrt(5/2))^2`
= `1/2 sqrt(2)/sqrt(5) tan^-1 ((sqrt(2)"t")/sqrt(5)) + "C"`
= `1/sqrt(10) tan^-1 ((sqrt(2)tanx)/sqrt(5)) + "C"`.
APPEARS IN
संबंधित प्रश्न
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
Find the integrals of the function:
sin2 (2x + 5)
Find the integrals of the function:
sin 3x cos 4x
Find the integrals of the function:
cos 2x cos 4x cos 6x
Find the integrals of the function:
sin3 x cos3 x
Find the integrals of the function:
sin x sin 2x sin 3x
Find the integrals of the function:
`cos x/(1 + cos x)`
Find the integrals of the function:
sin4 x
Find the integrals of the function:
`(cos 2x+ 2sin^2x)/(cos^2 x)`
`int (e^x(1 +x))/cos^2(e^x x) dx` equals ______.
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Evaluate `int_0^pi (x sin x)/(1 + cos^2 x) dx`
Evaluate `int_0^(3/2) |x sin pix|dx`
Find `int((3 sin x - 2) cos x)/(13 - cos^2 x- 7 sin x) dx`
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Find `int_ (sin2"x")/((sin^2 "x"+1)(sin^2"x"+3))d"x"`
Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Find: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Find: `int sin^-1 (2x) dx.`
Find `int x^2tan^-1x"d"x`
Evaluate the following:
`int tan^2x sec^4 x"d"x`
Evaluate the following:
`int (sinx + cosx)/sqrt(1 + sin 2x) "d"x`
Evaluate the following:
`int (sin^6x + cos^6x)/(sin^2x cos^2x) "d"x`
Evaluate the following:
`int "e"^(tan^-1x) ((1 + x + x^2)/(1 + x^2)) "d"x`
`int sinx/(3 + 4cos^2x) "d"x` = ______.
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
