What is the integral of 5sin(t/2)-4cos(2t) ?

The integral of 5sin(t/2)-4cos(2t) is -10cos(t/2)-2sin(2t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 5sin(t/2)-4cos(2t)

\int 5 sin (\frac{t}{2}) - 4 cos (2 t) d t

- 10 cos (\frac{t}{2}) - 2 sin (2 t) + C

Solution steps

\int 5 sin (\frac{t}{2}) - 4 cos (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 5 sin (\frac{t}{2}) d t - \int 4 cos (2 t) d t

\int 5 sin (\frac{t}{2}) d t = - 10 cos (\frac{t}{2})

\int 4 cos (2 t) d t = 2 sin (2 t)

= - 10 cos (\frac{t}{2}) - 2 sin (2 t)

Add a constant to the solution

= - 10 cos (\frac{t}{2}) - 2 sin (2 t) + C

\frac{d}{d x} ((5 x + 7)^{2})

\int_{0}^{0.3} 600 x d x

\int \frac{( 1 + 3 x ) ^{2}}{3 x} d x

\frac{d}{d x} (ln (\frac{1}{x}))

x \to 2 lim ((5 x + 7)^{4})

What is the integral of 5sin(t/2)-4cos(2t) ?
The integral of 5sin(t/2)-4cos(2t) is -10cos(t/2)-2sin(2t)+C