What is the integral of cos(8x+1) ?

The integral of cos(8x+1) is 1/8 sin(8x+1)+C

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integral of cos(8x+1)

\int cos (8 x + 1) d x

\frac{1}{8} sin (8 x + 1) + C

Solution steps

\int cos (8 x + 1) d x

Apply u-substitution

: \int cos (u) \frac{1}{8} d u

= \int cos (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{8} sin (u)

Substitute back

u = 8 x + 1

= \frac{1}{8} sin (8 x + 1)

Add a constant to the solution

= \frac{1}{8} sin (8 x + 1) + C

\int \frac{( x ^{8} + 1 )}{x ^{9} + 9 x} d x

\int 5 x (x^{2} + 2)^{2} d x

\int 8 x^{2} - 3 d x

\int x 4^{- x^{2}} d x

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

What is the integral of cos(8x+1) ?
The integral of cos(8x+1) is 1/8 sin(8x+1)+C