What is the integral of tan^2(9x) ?

The integral of tan^2(9x) is -x+1/9 tan(9x)+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(9x)

\int tan^{2} (9 x) d x

- x + \frac{1}{9} tan (9 x) + C

Solution steps

\int tan^{2} (9 x) d x

Rewrite using trig identities

= \int - 1 + sec^{2} (9 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 1 d x + \int sec^{2} (9 x) d x

\int 1 d x = x

\int sec^{2} (9 x) d x = \frac{1}{9} tan (9 x)

= - x + \frac{1}{9} tan (9 x)

Add a constant to the solution

= - x + \frac{1}{9} tan (9 x) + C

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

\int (L - x) d x

\int x sin (\frac{nπ x}{40}) d x

\int \frac{x ^{2}}{( x ^{2} + 8 ^{2} ) ^{\frac{3}{2}}} d x

\int (x + 3) (x - 6) d x

What is the integral of tan^2(9x) ?
The integral of tan^2(9x) is -x+1/9 tan(9x)+C