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

The integral of (sec(x)+3)^2 is tan(x)+6ln|tan(x)+sec(x)|+9x+C

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integral of (sec(x)+3)^2

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

tan (x) + 6 ln ∣ tan (x) + sec (x) ∣ + 9 x + C

Solution steps

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

Rewrite using trig identities

= \int \frac{( 1 + 3 cos ( x ) ) ^{2}}{cos ^{2} ( x )} d x

Expand

\frac{( 1 + 3 cos ( x ) ) ^{2}}{cos ^{2} ( x )} : \frac{1}{cos ^{2} ( x )} + \frac{6}{cos ( x )} + 9

= \int \frac{1}{cos ^{2} ( x )} + \frac{6}{cos ( x )} + 9 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{1}{cos ^{2} ( x )} d x + \int \frac{6}{cos ( x )} d x + \int 9 d x

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

\int \frac{6}{cos ( x )} d x = 6 ln ∣ tan (x) + sec (x) ∣

\int 9 d x = 9 x

= tan (x) + 6 ln ∣ tan (x) + sec (x) ∣ + 9 x

Add a constant to the solution

= tan (x) + 6 ln ∣ tan (x) + sec (x) ∣ + 9 x + C

\int (2 x e^{2 x}) d x

\int e^{7 x} d x

y^{^{''}} + 2 y^{^{'}} + 1 y = 1, y (1) = 1, y^{^{'}} (1) = 0

n \to \infty lim (n^{2} x)

f (x) = e^{13}

What is the integral of (sec(x)+3)^2 ?
The integral of (sec(x)+3)^2 is tan(x)+6ln|tan(x)+sec(x)|+9x+C