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

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

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

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

\frac{5}{2} x + \frac{1}{4} sin (2 x) + tan (x) + C

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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

\int 2 d x = 2 x

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

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

Simplify

\frac{1}{2} (x + \frac{1}{2} sin (2 x)) + 2 x + tan (x) : \frac{5}{2} x + \frac{1}{4} sin (2 x) + tan (x)

= \frac{5}{2} x + \frac{1}{4} sin (2 x) + tan (x)

Add a constant to the solution

= \frac{5}{2} x + \frac{1}{4} sin (2 x) + tan (x) + C

\int (4 x^{2} + 3 + \frac{5}{x ^{2} + 1}) d x

\int_{3}^{4} 5 x d x

y^{^{''}} + 9 y = 18

n = 1 \sum \infty \frac{15 n !}{n ^{n}}

\frac{\partial}{\partial x} (2 x + 3 y + 6 x y e^{3 x^{2} y})

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