What is the integral of cos(4x-3) ?

The integral of cos(4x-3) is 1/4 sin(4x-3)+C

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integral of cos(4x-3)

\int cos (4 x - 3) d x

\frac{1}{4} sin (4 x - 3) + C

Solution steps

\int cos (4 x - 3) d x

Apply u-substitution

: \int cos (u) \frac{1}{4} d u

= \int cos (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{4} sin (u)

Substitute back

u = 4 x - 3

= \frac{1}{4} sin (4 x - 3)

Add a constant to the solution

= \frac{1}{4} sin (4 x - 3) + C

n = 1 \sum \infty π^{n}

y = 3 x - x^{2}, y = - 3 x

\frac{\partial}{\partial x} (2^{2} + 1^{2})

(\frac{2}{7}, 5), (10, 39)

\int_{15}^{35} (x^{2} x) d x

What is the integral of cos(4x-3) ?
The integral of cos(4x-3) is 1/4 sin(4x-3)+C