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

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

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integral of (2tan(x))/(1-tan^2(x))

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

- \frac{1}{2} ln ∣ cos (2 x) ∣ + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply Integration By Parts

: x cos (x) sin (x) sec (2 x) - \int x sec^{2} (2 x) d x

= 2 (x cos (x) sin (x) sec (2 x) - \int x sec^{2} (2 x) d x)

\int x sec^{2} (2 x) d x = \frac{1}{2} x tan (2 x) + \frac{1}{4} ln ∣ cos (2 x) ∣

= 2 (x cos (x) sin (x) sec (2 x) - (\frac{1}{2} x tan (2 x) + \frac{1}{4} ln ∣ cos (2 x) ∣))

Simplify

2 (x cos (x) sin (x) sec (2 x) - (\frac{1}{2} x tan (2 x) + \frac{1}{4} ln ∣ cos (2 x) ∣)) : - \frac{1}{2} ln ∣ cos (2 x) ∣

= - \frac{1}{2} ln ∣ cos (2 x) ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ cos (2 x) ∣ + C

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

y = x^{3} + 6 x

4 + x^{6}

f (x) = 2 x 4 - x^{2}

x \to 2 lim (- x^{2} + 6 x - 4)

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