What is the integral of (5sec(x)-cos(x))^2 ?

The integral of (5sec(x)-cos(x))^2 is 25tan(x)-19/2 x+1/4 sin(2x)+C

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integral of (5sec(x)-cos(x))^2

\int (5 sec (x) - cos (x))^{2} d x

25 tan (x) - \frac{19}{2} x + \frac{1}{4} sin (2 x) + C

Solution steps

\int (5 sec (x) - cos (x))^{2} d x

Rewrite using trig identities

= \int \frac{( 5 - cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x )} d x

Expand

\frac{( 5 - cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x )} : \frac{25}{cos ^{2} ( x )} - 10 + cos^{2} (x)

= \int \frac{25}{cos ^{2} ( x )} - 10 + 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{25}{cos ^{2} ( x )} d x - \int 10 d x + \int cos^{2} (x) d x

\int \frac{25}{cos ^{2} ( x )} d x = 25 tan (x)

\int 10 d x = 10 x

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

= 25 tan (x) - 10 x + \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Simplify

25 tan (x) - 10 x + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : 25 tan (x) - \frac{19}{2} x + \frac{1}{4} sin (2 x)

= 25 tan (x) - \frac{19}{2} x + \frac{1}{4} sin (2 x)

Add a constant to the solution

= 25 tan (x) - \frac{19}{2} x + \frac{1}{4} sin (2 x) + C

\int cot^{3} (4 x) sec^{2} (4 x) d x

\int (4 - x^{2}) d x

x \to 0 - lim (\frac{e ^{x}}{x ^{3}})

\frac{y}{4}

\frac{\partial}{\partial x} ((x - \frac{a}{y ^{3}}) (b - x))

What is the integral of (5sec(x)-cos(x))^2 ?
The integral of (5sec(x)-cos(x))^2 is 25tan(x)-19/2 x+1/4 sin(2x)+C