What is the integral of (sec(x)+cos(x))/(7cos(x)) ?

The integral of (sec(x)+cos(x))/(7cos(x)) is 1/7 (tan(x)+x)+C

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integral of (sec(x)+cos(x))/(7cos(x))

\int \frac{sec ( x ) + cos ( x )}{7 cos ( x )} d x

\frac{1}{7} (tan (x) + x) + C

Solution steps

\int \frac{sec ( x ) + cos ( x )}{7 cos ( x )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int \frac{sec ( x ) + cos ( x )}{cos ( x )} d x

Expand

\frac{sec ( x ) + cos ( x )}{cos ( x )} : \frac{sec ( x )}{cos ( x )} + 1

= \frac{1}{7} \cdot \int \frac{sec ( x )}{cos ( x )} + 1 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

= \frac{1}{7} (\int \frac{sec ( x )}{cos ( x )} d x + \int 1 d x)

\int \frac{sec ( x )}{cos ( x )} d x = tan (x)

\int 1 d x = x

= \frac{1}{7} (tan (x) + x)

Add a constant to the solution

= \frac{1}{7} (tan (x) + x) + C

\frac{d}{d x} (x^{4} \frac{1}{8} + \frac{1}{4 x ^{2}})

\frac{d}{d x} ((6 x + 1)^{2})

y^{^{''}} - 25 y = 0

y^{^{'}} + 27 y = \frac{1}{x ^{2}}

\int \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} d x

What is the integral of (sec(x)+cos(x))/(7cos(x)) ?
The integral of (sec(x)+cos(x))/(7cos(x)) is 1/7 (tan(x)+x)+C