Advertisements
Advertisements
प्रश्न
Find the integrals of the function:
sin2 (2x + 5)
Advertisements
उत्तर
Let `I = int sin^2 (2x + 5)` dx
....`[because sin^2 A = (1 - cos 2A)/2]`
`= 1/2 int [(1 - cos (4x + 10)]` dx
`= 1/2 int dx - 1/2 int cos (4x + 10)` dx
`= 1/2 [x - (sin (4x + 10))/4] + C`
Put 4x + 10 = t
4 dx = dt
`x/2 - 1/2 . 1/4 int cos t dt`
`= x/2 - 1/8 sin t + C`
`= x/2 - 1/8 sin (4x + 10) + C`
APPEARS IN
संबंधित प्रश्न
Evaluate : `intsin(x-a)/sin(x+a)dx`
Find the integrals of the function:
sin 4x sin 8x
Find the integrals of the function:
sin4 x
Find the integrals of the function:
cos4 2x
Find the integrals of the function:
`(sin^2 x)/(1 + cos x)`
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:
tan4x
Find the integrals of the function:
`(cos 2x)/(cos x + sin x)^2`
Find the integrals of the function:
`1/(cos(x - a) cos(x - b))`
`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((3 sin x - 2) cos x)/(13 - cos^2 x- 7 sin x) dx`
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
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"`
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: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Find: `int sin^-1 (2x) dx.`
Evaluate `int tan^8 x sec^4 x"d"x`
Find `int "dx"/(2sin^2x + 5cos^2x)`
`int "dx"/(sin^2x cos^2x)` is equal to ______.
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
The value of the integral `int_(1/3)^1 (x - x^3)^(1/3)/x^4 dx` is
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
