What is the integral of 1/(cos(x)sin(x)+cos^2(x)) ?

The integral of 1/(cos(x)sin(x)+cos^2(x)) is 1/2 ln|1-tan^2(x)|+1/2 ln|tan(2x)+sec(2x)|+C

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integral of 1/(cos(x)sin(x)+cos^2(x))

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

\frac{1}{2} ln 1 - tan^{2} (x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣ + C

Solution steps

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

Multiply by the conjugate of

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

= \int \frac{1}{\frac{- c o s ^{2} ( x ) c o s ( 2 x )}{c o s ( x ) s i n ( x ) - c o s ^{2} ( x )}} d x

Take the constant out:

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

= - \int \frac{1}{\frac{c o s ^{2} ( x ) c o s ( 2 x )}{c o s ( x ) s i n ( x ) - c o s ^{2} ( x )}} d x

Expand

\frac{1}{\frac{c o s ^{2} ( x ) c o s ( 2 x )}{c o s ( x ) s i n ( x ) - c o s ^{2} ( x )}} : \frac{sin ( x )}{cos ( x ) cos ( 2 x )} - \frac{1}{cos ( 2 x )}

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

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

\int \frac{1}{cos ( 2 x )} d x = \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣

= - (- \frac{1}{2} ln 1 - tan^{2} (x) - \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣)

Simplify

= \frac{1}{2} ln 1 - tan^{2} (x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣

Add a constant to the solution

= \frac{1}{2} ln 1 - tan^{2} (x) + \frac{1}{2} ln ∣ tan (2 x) + sec (2 x) ∣ + C

\int 8 x ln (7 x) d x

\frac{d}{d x} (\frac{sin ( x )}{x ^{2}})

\int_{y}^{x} \frac{e ^{t}}{t} d t

(32) (41.5)

\frac{\partial}{\partial x} (\frac{x}{x ^{4} - y ^{6}})

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