Given the limitations of this browser, I will use S to denote integral (the stretched S).
Int = S cos(x)*sin3(x) dx
Let t = sin(x) then dt = cos(x)dx
so Int = S t3dt = 1/4*t4 + c
Substituting back for x,
Int = sin4(x) / 4 + c
Given the limitations of this browser, I will use S to denote integral (the stretched S).
Int = S cos(x)*sin3(x) dx
Let t = sin(x) then dt = cos(x)dx
so Int = S t3dt = 1/4*t4 + c
Substituting back for x,
Int = sin4(x) / 4 + c
Given the limitations of this browser, I will use S to denote integral (the stretched S).
Int = S cos(x)*sin3(x) dx
Let t = sin(x) then dt = cos(x)dx
so Int = S t3dt = 1/4*t4 + c
Substituting back for x,
Int = sin4(x) / 4 + c
Given the limitations of this browser, I will use S to denote integral (the stretched S).
Int = S cos(x)*sin3(x) dx
Let t = sin(x) then dt = cos(x)dx
so Int = S t3dt = 1/4*t4 + c
Substituting back for x,
Int = sin4(x) / 4 + c
Given the limitations of this browser, I will use S to denote integral (the stretched S).
Int = S cos(x)*sin3(x) dx
Let t = sin(x) then dt = cos(x)dx
so Int = S t3dt = 1/4*t4 + c
Substituting back for x,
Int = sin4(x) / 4 + c
(3pi-7)/9 To verify this go to the link and enter integrate x cos(x)^3 from 0 to pi/2.
-cos x + Constant
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
Let y = sin 2x Then dy/dx = 2*cos 2x Also, when x = 0, y = 0 and when x = pi/6, y = sin(2*pi/6) = sin(pi/3) = sqrt(3)/2 Therefore, the integral becomes definite integral from 0 to √3/2 of 1/2*y3dy = difference between 1/2*(y4)/4 = (y4)/8 evaluated at √3/2 and 0. = (√3/2)4 /8 = 9/128 = 0.0703 approx.
sin2x + c
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
sin integral is -cos This is so because the derivative of cos x = -sin x
(3pi-7)/9 To verify this go to the link and enter integrate x cos(x)^3 from 0 to pi/2.
The integral of x cos(x) dx is cos(x) + x sin(x) + C
The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.
-cos(3x) + constant
The integral of cos 5x is 1/5 sin (5x)
I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3
-cos x + Constant
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
3