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

The integral of (9+tan^2(θ)) is 8θ+tan(θ)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (9+tan^2(θ))

\int (9 + tan^{2} (θ)) d θ

8 θ + tan (θ) + C

Solution steps

\int 9 + tan^{2} (θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 9 d θ + \int tan^{2} (θ) d θ

\int 9 d θ = 9 θ

\int tan^{2} (θ) d θ = - θ + tan (θ)

= 9 θ - θ + tan (θ)

Simplify

= 8 θ + tan (θ)

Add a constant to the solution

= 8 θ + tan (θ) + C

\int 6 x 4 + x^{2} d x

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

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

\int x tan (x) tan (x) d x

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

What is the integral of (9+tan^2(θ)) ?
The integral of (9+tan^2(θ)) is 8θ+tan(θ)+C