What is the integral of 2x*cos(1+3x^2) ?

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

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integral of 2x*cos(1+3x^2)

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

\frac{1}{3} sin (1 + 3 x^{2}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{cos ( u )}{6} d u

= 2 \cdot \int \frac{cos ( u )}{6} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{6} sin (u)

Substitute back

u = 1 + 3 x^{2}

= 2 \cdot \frac{1}{6} sin (1 + 3 x^{2})

Simplify

2 \cdot \frac{1}{6} sin (1 + 3 x^{2}) : \frac{1}{3} sin (1 + 3 x^{2})

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

Add a constant to the solution

= \frac{1}{3} sin (1 + 3 x^{2}) + C

\int \frac{( 7 x ^{3} - 8 x )}{x} d x

\frac{d}{d x} (\frac{5 x ^{2} + 4}{x + 3})

x \to - 1 lim (\frac{( 1 - x )}{x + 1})

y = 3 6 x^{2} + 1

- 3^{- 2} (tan (7 t))^{5}

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