What is the integral of (2x+4)cos(x^2+4x-6) ?

The integral of (2x+4)cos(x^2+4x-6) is sin(x^2+4x-6)+C

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integral of (2x+4)cos(x^2+4x-6)

\int (2 x + 4) cos (x^{2} + 4 x - 6) d x

sin (x^{2} + 4 x - 6) + C

Solution steps

\int (2 x + 4) cos (x^{2} + 4 x - 6) d x

(2 x + 4) cos (x^{2} + 4 x - 6) = 2 (x + 2) cos (x^{2} + 4 x - 6)

= \int 2 (x + 2) cos (x^{2} + 4 x - 6) d x

Take the constant out:

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

= 2 \cdot \int (x + 2) cos (x^{2} + 4 x - 6) d x

Apply u-substitution

: \int \frac{cos ( u )}{2} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = x^{2} + 4 x - 6

= 2 \cdot \frac{1}{2} sin (x^{2} + 4 x - 6)

Simplify

2 \cdot \frac{1}{2} sin (x^{2} + 4 x - 6) : sin (x^{2} + 4 x - 6)

= sin (x^{2} + 4 x - 6)

Add a constant to the solution

= sin (x^{2} + 4 x - 6) + C

\int x sin (nπ x) d x

\frac{d}{d x} (3 sin (\frac{1}{6}) x)

\int_{0}^{x} t (e^{t^{3}} - 1) d t

\int_{1}^{3} 2 x - 3 d x

x (8 - y) = \frac{d y}{d x}

What is the integral of (2x+4)cos(x^2+4x-6) ?
The integral of (2x+4)cos(x^2+4x-6) is sin(x^2+4x-6)+C