What is the integral of 3cos(1/2 x) ?

The integral of 3cos(1/2 x) is 6sin(1/2 x)+C

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integral of 3cos(1/2 x)

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

6 sin (\frac{1}{2} x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int cos (\frac{1}{2} x) d x

Apply u-substitution

: \int cos (u) \cdot 2 d u

= 3 \cdot \int cos (u) \cdot 2 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot 2 \cdot \int cos (u) d u

Use the common integral:

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

= 3 \cdot 2 sin (u)

Substitute back

u = \frac{1}{2} x

= 3 \cdot 2 sin (\frac{1}{2} x)

Simplify

= 6 sin (\frac{1}{2} x)

Add a constant to the solution

= 6 sin (\frac{1}{2} x) + C

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

n = 1 \sum \infty ln (\frac{n}{3 n + 1})

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

\int 6 - x^{2} d x

\int \frac{1}{1 - 2 x - x ^{2}} d x

What is the integral of 3cos(1/2 x) ?
The integral of 3cos(1/2 x) is 6sin(1/2 x)+C