sin2x + c
(1/8)(x-sin 4x)
- cos(1 - X) + C
∫ cos(x) dx = -sin(x) + C
If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration is a function of x, here x^2. The function being integrated is sin(t)
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
The integral of x cos(x) dx is cos(x) + x sin(x) + C
sin2x + c
(1/8)(x-sin 4x)
- cos(1 - X) + C
∫ cos(x) dx = -sin(x) + C
.5(x-sin(x)cos(x))+c
If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration is a function of x, here x^2. The function being integrated is sin(t)
∫ cot(x) dx is written as: ∫ cos(x) / sin(x) dx Let u = sin(x). Then, du = cos(x) dx, giving us: ∫ 1/u du So the integral of 1/u is ln|u|. So the answer is ln|sin(x)| + c
The method to use is 'integration by parts'; set u =x; du=dx; dv = sin(pi x)dx; v = cos(pi x)/pi. so integral(u dv) = u*v - integral(v du) then repeat the process.
The integral of cot(x)dx is ln|sin(x)| + C