What is the integral of (1+sin(x))^2 ?

The integral of (1+sin(x))^2 is 3/2 x-1/4 sin(2x)-2cos(x)+C

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integral of (1+sin(x))^2

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

\frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x) + C

Solution steps

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

Expand

(1 + sin (x))^{2} : 1 + 2 sin (x) + sin^{2} (x)

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

\int 1 d x = x

\int 2 sin (x) d x = - 2 cos (x)

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

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

Simplify

x - 2 cos (x) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : \frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x)

= \frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x)

Add a constant to the solution

= \frac{3}{2} x - \frac{1}{4} sin (2 x) - 2 cos (x) + C

(x^{2} + y^{2} + x) d x + x y d y = 0

\int \frac{y}{y ^{3} - 1} d y

\int x e^{- 2 y} d x

\frac{\partial}{\partial x} (x \cdot e^{\frac{y}{z}})

n = 1 \sum \infty \frac{n}{n ^{5} + 1}

What is the integral of (1+sin(x))^2 ?
The integral of (1+sin(x))^2 is 3/2 x-1/4 sin(2x)-2cos(x)+C