What is the integral of (rivative of cos(2x))^2 ?

The integral of (rivative of cos(2x))^2 is derivative of cos(2x)^2x-8(-1/4 xcos(4x)+1/16 sin(4x))+C

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integral of (rivative of cos(2x))^2

\int (\frac{d}{d x} (cos (2 x)))^{2} d x

\frac{d}{d x} (cos (2 x))^{2} x - 8 (- \frac{1}{4} x cos (4 x) + \frac{1}{16} sin (4 x)) + C

Solution steps

\int \frac{d}{d x} (cos (2 x))^{2} d x

Apply Integration By Parts

: \frac{d}{d x} (cos (2 x))^{2} x - \int 8 sin (4 x) x d x

= \frac{d}{d x} (cos (2 x))^{2} x - \int 8 sin (4 x) x d x

\int 8 sin (4 x) x d x = 8 (- \frac{1}{4} x cos (4 x) + \frac{1}{16} sin (4 x))

= \frac{d}{d x} (cos (2 x))^{2} x - 8 (- \frac{1}{4} x cos (4 x) + \frac{1}{16} sin (4 x))

Add a constant to the solution

= \frac{d}{d x} (cos (2 x))^{2} x - 8 (- \frac{1}{4} x cos (4 x) + \frac{1}{16} sin (4 x)) + C

x - x e^{5 x + 2}

m x^{2} - p x^{2} + p a^{2} - m a^{2}

3^{12} \cdot 7

x + 3 - 4

f (x) (x^{2} - 1)^{l n (x^{2} - 1)}

What is the integral of (rivative of cos(2x))^2 ?
The integral of (rivative of cos(2x))^2 is derivative of cos(2x)^2x-8(-1/4 xcos(4x)+1/16 sin(4x))+C