Advertisements
Advertisements
Question
Find the integrals of the function:
cos4 2x
Solution
Let `I = int cos^4 2x dx`
`= int ((1 + cos 4x)/2)^2 dx`
`= 1/4 int (1 + cos^2 4x + 2 cos 4x) dx`
`= 1/4 int [1 + (1 + cos 8x)/2 + 2 cos 4x] dx`
`= 3/8 int dx + 1/8 int cos 8x dx + 1/2 int cos 4x dx`
`= 3/8 x + 1/64 sin 8x + 1/8 sin 4x + C`
APPEARS IN
RELATED QUESTIONS
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
Find the integrals of the function:
cos 2x cos 4x cos 6x
Find the integrals of the function:
sin 4x sin 8x
Find the integrals of the function:
tan4x
Find the integrals of the function:
`1/(sin xcos^3 x)`
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 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 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: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
Find: `int sin^-1 (2x) dx.`
`int "e"^x (cosx - sinx)"d"x` is equal to ______.
`int "dx"/(sin^2x cos^2x)` is equal to ______.
`int (sin^6x)/(cos^8x) "d"x` = ______.
Evaluate the following:
`int ((1 + cosx))/(x + sinx) "d"x`
Evaluate the following:
`int ("d"x)/(1 + cos x)`
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
Evaluate the following:
`int (sin^6x + cos^6x)/(sin^2x cos^2x) "d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
Evaluate the following:
`int "e"^(tan^-1x) ((1 + x + x^2)/(1 + x^2)) "d"x`
Evaluate the following:
`int sin^-1 sqrt(x/("a" + x)) "d"x` (Hint: Put x = a tan2θ)
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
`int (cos^2x)/(sin x + cos x)^2 dx` is equal to
