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

The integral of (3sec(x)-cos(x))^2 is 9tan(x)-11/2 x+1/4 sin(2x)+C

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

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

9 tan (x) - \frac{11}{2} x + \frac{1}{4} sin (2 x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{( 3 - cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x )} d x

Expand

\frac{( 3 - cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x )} : \frac{9}{cos ^{2} ( x )} - 6 + cos^{2} (x)

= \int \frac{9}{cos ^{2} ( x )} - 6 + 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{9}{cos ^{2} ( x )} d x - \int 6 d x + \int cos^{2} (x) d x

\int \frac{9}{cos ^{2} ( x )} d x = 9 tan (x)

\int 6 d x = 6 x

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

= 9 tan (x) - 6 x + \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Simplify

9 tan (x) - 6 x + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : 9 tan (x) - \frac{11}{2} x + \frac{1}{4} sin (2 x)

= 9 tan (x) - \frac{11}{2} x + \frac{1}{4} sin (2 x)

Add a constant to the solution

= 9 tan (x) - \frac{11}{2} x + \frac{1}{4} sin (2 x) + C

x^{2} - 4 x + 39

\int_{1}^{4} π (\frac{1}{x})^{2} d x

y^{^{'}} = 3 x^{2} y

x \to 0 + lim (\frac{ln ( x )}{csc ( x )})

f (x) = \frac{1}{( 1 - x ) ^{2}}

What is the integral of (3sec(x)-cos(x))^2 ?
The integral of (3sec(x)-cos(x))^2 is 9tan(x)-11/2 x+1/4 sin(2x)+C