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

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

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integral of-sec^2(x)+cos(2x)

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

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

Solution steps

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

\int sec^{2} (x) d x = tan (x)

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

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

Add a constant to the solution

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

\frac{( 5 x - 2 )}{( x ^{2} + 1 )}

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

\frac{2 u ^{2}}{( u ^{2} + u ) ^{3}}

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

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

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