What is the integral of-4sin(x)+9cos(x) ?

The integral of-4sin(x)+9cos(x) is 4cos(x)+9sin(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-4sin(x)+9cos(x)

\int - 4 sin (x) + 9 cos (x) d x

4 cos (x) + 9 sin (x) + C

Solution steps

\int - 4 sin (x) + 9 cos (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 4 sin (x) d x + \int 9 cos (x) d x

\int 4 sin (x) d x = - 4 cos (x)

\int 9 cos (x) d x = 9 sin (x)

= - (- 4 cos (x)) + 9 sin (x)

Simplify

= 4 cos (x) + 9 sin (x)

Add a constant to the solution

= 4 cos (x) + 9 sin (x) + C

\int cot (\frac{5}{8}) x \frac{5}{8} d x

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

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

\int (x + \frac{sin ( x )}{x}) d x

cos (π x)

What is the integral of-4sin(x)+9cos(x) ?
The integral of-4sin(x)+9cos(x) is 4cos(x)+9sin(x)+C