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

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

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

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

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

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{tan ^{2} ( x )}{sec ( x )} d x

Rewrite using trig identities

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

Expand

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

= \frac{1}{2} \cdot \int - \frac{1}{sec ( x )} + sec (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

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

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

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

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

Add a constant to the solution

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

\int_{0}^{π} 4 x cos (x) d x

\frac{d}{d x} (\frac{1}{in ( x - 1 )})

(e^{\frac{- x ^{2}}{2}})^{^{'}}

\int \frac{3}{1 - x} d x

\frac{d}{d x} (x^{2} - 4 x + 3)

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