What is the integral of [3sin(x)+(1/2)tan(x)] ?

The integral of [3sin(x)+(1/2)tan(x)] is -3cos(x)-1/2 ln|cos(x)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of [3sin(x)+(1/2)tan(x)]

\int [3 sin (x) + (\frac{1}{2}) tan (x)] d x

- 3 cos (x) - \frac{1}{2} ln ∣ cos (x) ∣ + C

Solution steps

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

\int 3 sin (x) d x = - 3 cos (x)

\int \frac{1}{2} tan (x) d x = - \frac{1}{2} ln ∣ cos (x) ∣

= - 3 cos (x) - \frac{1}{2} ln ∣ cos (x) ∣

Add a constant to the solution

= - 3 cos (x) - \frac{1}{2} ln ∣ cos (x) ∣ + C

f (x) = 6 - 5 x^{2}

4 x^{7}

y = x^{2} + 3 x^{2} ln (x)

\int \frac{e ^{2 x} - 7}{e ^{2 x}} d x

\frac{7}{8 t ^{4}}

What is the integral of [3sin(x)+(1/2)tan(x)] ?
The integral of [3sin(x)+(1/2)tan(x)] is -3cos(x)-1/2 ln|cos(x)|+C