Advertisements
Advertisements
प्रश्न
Find `int "dx"/(2sin^2x + 5cos^2x)`
Advertisements
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:
sin3 (2x + 1)
Find the integrals of the function:
sin 4x sin 8x
Find the integrals of the function:
`(1-cosx)/(1 + cos x)`
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:
tan3 2x sec 2x
Find the integrals of the function:
tan4x
Find the integrals of the function:
`(sin^3 x + cos^3 x)/(sin^2x cos^2 x)`
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
Find the integrals of the function:
sin−1 (cos x)
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Find `int dx/(x^2 + 4x + 8)`
Find `int (2x)/((x^2 + 1)(x^4 + 4))`dx
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
Find `int_ sin ("x" - a)/(sin ("x" + a )) d"x"`
Find `int_ (log "x")^2 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:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
`int "dx"/(sin^2x cos^2x)` is equal to ______.
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Evaluate the following:
`int tan^2x sec^4 x"d"x`
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
Evaluate the following:
`int "e"^(tan^-1x) ((1 + x + x^2)/(1 + x^2)) "d"x`
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
