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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

= \int \frac{2 ( u ^{2} + 1 )}{( - u ^{2} + 2 u + 1 ) ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2} + 1}{( - u ^{2} + 2 u + 1 ) ( - u ^{2} + 1 )} d u

Take the partial fraction of

\frac{u ^{2} + 1}{( - u ^{2} + 2 u + 1 ) ( - u ^{2} + 1 )} : \frac{u - 1}{u ^{2} - 2 u - 1} - \frac{1}{2 ( u + 1 )} - \frac{1}{2 ( u - 1 )}

= 2 \cdot \int \frac{u - 1}{u ^{2} - 2 u - 1} - \frac{1}{2 ( u + 1 )} - \frac{1}{2 ( u - 1 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{u - 1}{u ^{2} - 2 u - 1} d u - \int \frac{1}{2 ( u + 1 )} d u - \int \frac{1}{2 ( u - 1 )} d u)

\int \frac{u - 1}{u ^{2} - 2 u - 1} d u = \frac{1}{2} ln u^{2} - 2 u - 1

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

\int \frac{1}{2 ( u - 1 )} d u = \frac{1}{2} ln ∣ u - 1 ∣

= 2 (\frac{1}{2} ln u^{2} - 2 u - 1 - \frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

Substitute back

u = tan (\frac{x}{2})

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

Simplify

2 (\frac{1}{2} ln tan^{2} (\frac{x}{2}) - 2 tan (\frac{x}{2}) - 1 - \frac{1}{2} ln tan (\frac{x}{2}) + 1 - \frac{1}{2} ln tan (\frac{x}{2}) - 1) : ln tan^{2} (\frac{x}{2}) - 2 tan (\frac{x}{2}) - 1 - ln tan (\frac{x}{2}) + 1 - ln tan (\frac{x}{2}) - 1

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

Add a constant to the solution

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

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