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)

integral

1 + sin^{2} (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

\int 1 + sin (x) cos (x) d x

\int x + \frac{2}{sin ( x )} d x

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

\int (x + 1)^{3} (x - 2) d 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