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

The integral of (sin^2(x)+cos(x))^2 is 7/8 x-4/3 sin^3(x)+1/32 sin(4x)+sin(x)-cos(2x)sin(x)+C

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

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

\frac{7}{8} x - \frac{4}{3} sin^{3} (x) + \frac{1}{32} sin (4 x) + sin (x) - cos (2 x) sin (x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{4} (1 - cos (2 x))^{2} + cos (x) - cos (x) cos (2 x) + cos^{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 \frac{1}{4} (1 - cos (2 x))^{2} d x + \int cos (x) d x - \int cos (x) cos (2 x) d x + \int cos^{2} (x) d x

\int \frac{1}{4} (1 - cos (2 x))^{2} d x = \frac{1}{32} (12 x - 8 sin (2 x) + sin (4 x))

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

\int cos (x) cos (2 x) d x = cos (2 x) sin (x) + \frac{4}{3} sin^{3} (x)

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

= \frac{1}{32} (12 x - 8 sin (2 x) + sin (4 x)) + sin (x) - (cos (2 x) sin (x) + \frac{4}{3} sin^{3} (x)) + \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Simplify

\frac{1}{32} (12 x - 8 sin (2 x) + sin (4 x)) + sin (x) - (cos (2 x) sin (x) + \frac{4}{3} sin^{3} (x)) + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : \frac{7}{8} x - \frac{4}{3} sin^{3} (x) + \frac{1}{32} sin (4 x) + sin (x) - cos (2 x) sin (x)

= \frac{7}{8} x - \frac{4}{3} sin^{3} (x) + \frac{1}{32} sin (4 x) + sin (x) - cos (2 x) sin (x)

Add a constant to the solution

= \frac{7}{8} x - \frac{4}{3} sin^{3} (x) + \frac{1}{32} sin (4 x) + sin (x) - cos (2 x) sin (x) + C

f (x) = x - 2 sin (x)

\int_{0}^{2} \frac{x}{x ^{4} + 1} d x

e^{- x^{2}}, 0

\frac{d}{d x} ((x^{4} + 1)^{\frac{1}{2}})

\int 2 z d z

What is the integral of (sin^2(x)+cos(x))^2 ?
The integral of (sin^2(x)+cos(x))^2 is 7/8 x-4/3 sin^3(x)+1/32 sin(4x)+sin(x)-cos(2x)sin(x)+C