What is the integral of (2t^2+sec^2(t)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2t^2+sec^2(t))

\int (2 t^{2} + sec^{2} (t)) d t

\frac{2 t ^{3}}{3} + tan (t) + C

Solution steps

\int 2 t^{2} + sec^{2} (t) d t

Apply the Sum Rule:

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

= \int 2 t^{2} d t + \int sec^{2} (t) d t

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

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

= \frac{2 t ^{3}}{3} + tan (t)

Add a constant to the solution

= \frac{2 t ^{3}}{3} + tan (t) + C

\int x^{2} (20 x^{7} - 7 x^{4} + 6) d x

sin (x^{5})

x = 7 y^{2}, x = 4 + 3 y^{2}

\int x ln (x) - x d x

\int \frac{1}{2 x ^{2} - 5 x + 3} d x

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