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

The integral of tan^2(arctan(x) is -x+1/(arctan(d))tan(xarctan(d))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^2(arctan(x)

\int tan^{2} (arctan (d) x) d x

- x + \frac{1}{arctan ( d )} tan (x arctan (d)) + C

Solution steps

\int tan^{2} (arctan (d) x) d x

Apply u-substitution

: \int \frac{tan ^{2} ( u )}{arctan ( d )} d u

= \int \frac{tan ^{2} ( u )}{arctan ( d )} d u

Take the constant out:

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

= \frac{1}{arctan ( d )} \cdot \int tan^{2} (u) d u

Rewrite using trig identities

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

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

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= \frac{1}{arctan ( d )} (- u + tan (u))

Substitute back

u = arctan (d) x

= \frac{1}{arctan ( d )} (- arctan (d) x + tan (arctan (d) x))

Simplify

\frac{1}{arctan ( d )} (- arctan (d) x + tan (arctan (d) x)) : - x + \frac{1}{arctan ( d )} tan (x arctan (d))

= - x + \frac{1}{arctan ( d )} tan (x arctan (d))

Add a constant to the solution

= - x + \frac{1}{arctan ( d )} tan (x arctan (d)) + C

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

\int \frac{1}{4 + 2 x} d x

\int \frac{z + 5}{z ( z ^{2} + 49 )} d z

\int \frac{w}{1 - w ^{2}} d w

\int 14 t e^{t} d t

What is the integral of tan^2(arctan(x) ?
The integral of tan^2(arctan(x) is -x+1/(arctan(d))tan(xarctan(d))+C