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

The integral of (1+sin^2(x))(1+cos^2(x)) is 17/8 x-1/32 sin(4x)+C

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

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

\frac{17}{8} x - \frac{1}{32} sin (4 x) + C

Solution steps

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

Expand

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

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

\int 1 d x = x

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

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

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

= x + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{8} (x - \frac{1}{4} sin (4 x))

Simplify

x + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : \frac{17}{8} x - \frac{1}{32} sin (4 x)

= \frac{17}{8} x - \frac{1}{32} sin (4 x)

Add a constant to the solution

= \frac{17}{8} x - \frac{1}{32} sin (4 x) + C

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What is the integral of (1+sin^2(x))(1+cos^2(x)) ?
The integral of (1+sin^2(x))(1+cos^2(x)) is 17/8 x-1/32 sin(4x)+C