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

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

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

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

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

Solution steps

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

\int 1 d x = x

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln ((x - 4)^{3}))

\int x \cdot x^{2} + 5 d x

ln (4 - x)

\frac{\partial}{\partial z} (ln (1 + x^{2} + e^{y} + z))

x \to 0 lim (\frac{1}{1 - e ^{- x}})

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