What is the integral of (7sec(x)-cos(x))^2 ?

The integral of (7sec(x)-cos(x))^2 is 49tan(x)-27/2 x+1/4 sin(2x)+C

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integral of (7sec(x)-cos(x))^2

\int (7 sec (x) - cos (x))^{2} d x

49 tan (x) - \frac{27}{2} x + \frac{1}{4} sin (2 x) + C

Solution steps

\int (7 sec (x) - cos (x))^{2} d x

Rewrite using trig identities

= \int \frac{( 7 - cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x )} d x

Expand

\frac{( 7 - cos ^{2} ( x ) ) ^{2}}{cos ^{2} ( x )} : \frac{49}{cos ^{2} ( x )} - 14 + cos^{2} (x)

= \int \frac{49}{cos ^{2} ( x )} - 14 + 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{49}{cos ^{2} ( x )} d x - \int 14 d x + \int cos^{2} (x) d x

\int \frac{49}{cos ^{2} ( x )} d x = 49 tan (x)

\int 14 d x = 14 x

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

= 49 tan (x) - 14 x + \frac{1}{2} (x + \frac{1}{2} sin (2 x))

Simplify

49 tan (x) - 14 x + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : 49 tan (x) - \frac{27}{2} x + \frac{1}{4} sin (2 x)

= 49 tan (x) - \frac{27}{2} x + \frac{1}{4} sin (2 x)

Add a constant to the solution

= 49 tan (x) - \frac{27}{2} x + \frac{1}{4} sin (2 x) + C

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What is the integral of (7sec(x)-cos(x))^2 ?
The integral of (7sec(x)-cos(x))^2 is 49tan(x)-27/2 x+1/4 sin(2x)+C