What is the integral of sin^2(2x+5) ?

The integral of sin^2(2x+5) is 1/8 (2(2x+5)-sin(2(2x+5)))+C

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integral of sin^2(2x+5)

\int sin^{2} (2 x + 5) d x

\frac{1}{8} (2 (2 x + 5) - sin (2 (2 x + 5))) + C

Solution steps

\int sin^{2} (2 x + 5) d x

Rewrite using trig identities

= \int \frac{1 - cos ( 2 ( 2 x + 5 ) )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 1 - cos (2 (2 x + 5)) d x

Apply u-substitution

: \int (1 - cos (u)) \frac{1}{4} d u

= \frac{1}{2} \cdot \int (1 - cos (u)) \frac{1}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int 1 - cos (u) d u

Apply the Sum Rule:

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

= \frac{1}{2} \cdot \frac{1}{4} (\int 1 d u - \int cos (u) d u)

\int 1 d u = u

\int cos (u) d u = sin (u)

= \frac{1}{2} \cdot \frac{1}{4} (u - sin (u))

Substitute back

u = 2 (2 x + 5)

= \frac{1}{2} \cdot \frac{1}{4} (2 (2 x + 5) - sin (2 (2 x + 5)))

Simplify

\frac{1}{2} \cdot \frac{1}{4} (2 (2 x + 5) - sin (2 (2 x + 5))) : \frac{1}{8} (2 (2 x + 5) - sin (2 (2 x + 5)))

= \frac{1}{8} (2 (2 x + 5) - sin (2 (2 x + 5)))

Add a constant to the solution

= \frac{1}{8} (2 (2 x + 5) - sin (2 (2 x + 5))) + C

\frac{\partial}{\partial x} (e^{- 7 y} cos (7 x))

\frac{d y}{d x} = \frac{y}{x} + \frac{x}{2 ( y + x )}

\int_{0}^{1} (2 x + 1)^{3} d x

f (x) = 3 x + 7, (3, 4)

\frac{1}{s ( s ^{2} + 2 s + 2 )}

What is the integral of sin^2(2x+5) ?
The integral of sin^2(2x+5) is 1/8 (2(2x+5)-sin(2(2x+5)))+C