What is the integral of (sin(2x)+cos(2x)) ?

The integral of (sin(2x)+cos(2x)) is 1/2 sin(2x)-1/2 cos(2x)+C

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integral of (sin(2x)+cos(2x))

\int (sin (2 x) + cos (2 x)) d x

\frac{1}{2} sin (2 x) - \frac{1}{2} cos (2 x) + C

Solution steps

\int sin (2 x) + cos (2 x) d x

Rewrite using trig identities

= \int - 1 + 2 cos^{2} (x) + sin (2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d x + \int 2 cos^{2} (x) d x + \int sin (2 x) d x

\int 1 d x = x

\int 2 cos^{2} (x) d x = x + \frac{1}{2} sin (2 x)

\int sin (2 x) d x = - \frac{1}{2} cos (2 x)

= - x + x + \frac{1}{2} sin (2 x) - \frac{1}{2} cos (2 x)

Simplify

= \frac{1}{2} sin (2 x) - \frac{1}{2} cos (2 x)

Add a constant to the solution

= \frac{1}{2} sin (2 x) - \frac{1}{2} cos (2 x) + C

y = x^{2}, y = 2 x - x^{2}

\frac{d}{d x} (\frac{q x + r ( q , x , t )}{s x + t})

\int (x + 4) 6 - x d x

\int \frac{2 x ^{3}}{x ^{2} - 4} d x

\frac{d}{d x} (3 x^{3} + 2 x + 9)

What is the integral of (sin(2x)+cos(2x)) ?
The integral of (sin(2x)+cos(2x)) is 1/2 sin(2x)-1/2 cos(2x)+C