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

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

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

integral

\frac{tan ^{2} ( x )}{cos ( x )}

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

Solution steps

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

Rewrite using trig identities

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

\frac{- 1 + sec ^{2} ( x )}{cos ( x )} = - \frac{1}{cos ( x )} + \frac{sec ^{2} ( x )}{cos ( x )}

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

Rewrite using trig identities

= \int - sec (x) + sec^{3} (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 sec (x) d x + \int sec^{3} (x) d x

\int sec (x) d x = ln ∣ tan (x) + sec (x) ∣

\int sec^{3} (x) d x = \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

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

Simplify

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

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

Add a constant to the solution

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

\frac{( 4 - x )}{x}

\frac{( e ^{- t} - 4 )}{e ^{- t}}

e^{6} x cos (e^{3} x)

2 x + 5

\frac{1}{( x ^{3} + a ^{3} )}

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