What is the integral of-3cos(x)+4sec^2(x) ?

The integral of-3cos(x)+4sec^2(x) is -3sin(x)+4tan(x)+C

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integral of-3cos(x)+4sec^2(x)

\int - 3 cos (x) + 4 sec^{2} (x) d x

- 3 sin (x) + 4 tan (x) + C

Solution steps

\int - 3 cos (x) + 4 sec^{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 3 cos (x) d x + \int 4 sec^{2} (x) d x

\int 3 cos (x) d x = 3 sin (x)

\int 4 sec^{2} (x) d x = 4 tan (x)

= - 3 sin (x) + 4 tan (x)

Add a constant to the solution

= - 3 sin (x) + 4 tan (x) + C

\frac{\partial}{\partial x} (4 cos (4 x))

\frac{1}{x + 1}

\int (t + 4) e^{t^{2} + 8 t + 1} d t

y^{^{''}} + 8 y^{^{'}} = - 256 cos (8 x)

y^{^{'}} + 2 x^{- 1} y = 6 x^{4} y^{2}

What is the integral of-3cos(x)+4sec^2(x) ?
The integral of-3cos(x)+4sec^2(x) is -3sin(x)+4tan(x)+C