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

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

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

Multiply by the conjugate of

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

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

Expand

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

n = 0 \sum \infty \frac{1}{( 4 n ^{2} - 1 )}

\int_{8}^{\infty} x e^{- 2 x} d x

\frac{d}{d x} (sin^{3} (x) + sin (3 x))

\frac{d}{d y} (tan (x) + y)

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