Advertisements
Advertisements
Question
Find the integrals of the function:
`(cos 2x+ 2sin^2x)/(cos^2 x)`
Solution
Let `I = int (cos 2x + 2 sin^2)/cos^2 x dx`
`= int ((cos^2 x - sin^2 x) + 2 sin^2 x)/cos^2 x` .... [∵ cos 2x = cos2 x - sin2 x]
`= int(cos^2x + sin^2x)/cos^2 x dx`
`= 1/cos^2 x dx`
`= int sec^2 x dx = tan x + C`
APPEARS IN
RELATED QUESTIONS
Evaluate : `intsin(x-a)/sin(x+a)dx`
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 x sin 2x sin 3x
Find the integrals of the function:
sin 4x sin 8x
Find the integrals of the function:
cos4 2x
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
`(cos x - sinx)/(1+sin 2x)`
Find the integrals of the function:
tan3 2x sec 2x
Find the integrals of the function:
tan4x
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
`int (sin^2x - cos^2 x)/(sin^2 x cos^2 x) dx` is equal to ______.
`int (e^x(1 +x))/cos^2(e^x x) dx` equals ______.
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
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
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"`
Integrate the function `cos("x + a")/sin("x + b")` w.r.t. x.
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
Find: `int sin^-1 (2x) dx.`
Evaluate `int tan^8 x sec^4 x"d"x`
Find `int x^2tan^-1x"d"x`
`int "dx"/(sin^2x cos^2x)` is equal to ______.
Evaluate the following:
`int ((1 + cosx))/(x + sinx) "d"x`
Evaluate the following:
`int (sinx + cosx)/sqrt(1 + sin 2x) "d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
`int sinx/(3 + 4cos^2x) "d"x` = ______.
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to