- cos(1 - X) + C
-cos x + Constant
To find the integral of ( \sin^2(2x) \cos^2(2x) ), you can use the identity ( \sin^2(a) \cos^2(a) = \frac{1}{4} \sin^2(2a) ). Thus, we have: [ \sin^2(2x) \cos^2(2x) = \frac{1}{4} \sin^2(4x). ] The integral then becomes: [ \int \sin^2(2x) \cos^2(2x) , dx = \frac{1}{4} \int \sin^2(4x) , dx. ] Using the identity ( \sin^2(a) = \frac{1 - \cos(2a)}{2} ), you can further simplify and evaluate the integral.
sin2x + c
The indefinite integral of sin 2x is -cos 2x / 2 + C, where C is any constant.
(1/8)(x-sin 4x)
0.5 x2+C
sin integral is -cos This is so because the derivative of cos x = -sin x
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
-cos x + Constant
To find the integral of ( \sin^2(2x) \cos^2(2x) ), you can use the identity ( \sin^2(a) \cos^2(a) = \frac{1}{4} \sin^2(2a) ). Thus, we have: [ \sin^2(2x) \cos^2(2x) = \frac{1}{4} \sin^2(4x). ] The integral then becomes: [ \int \sin^2(2x) \cos^2(2x) , dx = \frac{1}{4} \int \sin^2(4x) , dx. ] Using the identity ( \sin^2(a) = \frac{1 - \cos(2a)}{2} ), you can further simplify and evaluate the integral.
sin2x + c
The integral of cosine cubed is sinx- 1/3 sin cubed x + c
The indefinite integral of sin 2x is -cos 2x / 2 + C, where C is any constant.
(1/8)(x-sin 4x)
(5.4 / k) cos(kt)
-cos(x) + constant
∫ cos(x) dx = -sin(x) + C