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

The integral of (1+tan^2(x))sec^2(x) is tan(x)+(tan^3(x))/3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1+tan^2(x))sec^2(x)

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

tan (x) + \frac{tan ^{3} ( x )}{3} + C

Solution steps

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

Apply u-substitution

: \int 1 + u^{2} d u

= \int 1 + u^{2} 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

= \int 1 d u + \int u^{2} d u

\int 1 d u = u

\int u^{2} d u = \frac{u ^{3}}{3}

= u + \frac{u ^{3}}{3}

Substitute back

u = tan (x)

= tan (x) + \frac{tan ^{3} ( x )}{3}

Add a constant to the solution

= tan (x) + \frac{tan ^{3} ( x )}{3} + C

\int (3 - 2 x) d x

\int 4 x^{7} d x

\int 4 t e^{2 t} d t

\int 20 x^{3} (1 - x) d x

\int cos (2 x) e^{x} d x

What is the integral of (1+tan^2(x))sec^2(x) ?
The integral of (1+tan^2(x))sec^2(x) is tan(x)+(tan^3(x))/3+C