What is the integral of sin(t)1+cos(t) ?

The integral of sin(t)1+cos(t) is -cos(t)+sin(t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(t)1+cos(t)

\int sin (t) 1 + cos (t) d t

- cos (t) + sin (t) + C

Solution steps

\int sin (t) \cdot 1 + cos (t) d t

Simplify

= \int sin (t) + cos (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 sin (t) d t + \int cos (t) d t

\int sin (t) d t = - cos (t)

\int cos (t) d t = sin (t)

= - cos (t) + sin (t)

Add a constant to the solution

= - cos (t) + sin (t) + C

f (x) = \frac{x}{x + 5}, at x = 4

(3, 5), (- 1, 5)

\int_{0}^{4} x x d x

\frac{d}{d x} (4 e^{x} sin (x))

\frac{d}{d x} (2 - y - x^{2})

What is the integral of sin(t)1+cos(t) ?
The integral of sin(t)1+cos(t) is -cos(t)+sin(t)+C