What is the integral of cos^2(ax+b) ?

The integral of cos^2(ax+b) is x/2+b/(2a)+1/(4a)sin(2(ax+b))+C

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integral of cos^2(ax+b)

\int cos^{2} (a x + b) d x

\frac{x}{2} + \frac{b}{2 a} + \frac{1}{4 a} sin (2 (a x + b)) + C

Solution steps

\int cos^{2} (a x + b) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

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

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

Substitute back

u = a x + b

= \frac{1}{a} \cdot \frac{1}{2} (a x + b + \frac{1}{2} sin (2 (a x + b)))

Simplify

\frac{1}{a} \cdot \frac{1}{2} (a x + b + \frac{1}{2} sin (2 (a x + b))) : \frac{x}{2} + \frac{b}{2 a} + \frac{1}{4 a} sin (2 (a x + b))

= \frac{x}{2} + \frac{b}{2 a} + \frac{1}{4 a} sin (2 (a x + b))

Add a constant to the solution

= \frac{x}{2} + \frac{b}{2 a} + \frac{1}{4 a} sin (2 (a x + b)) + C

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What is the integral of cos^2(ax+b) ?
The integral of cos^2(ax+b) is x/2+b/(2a)+1/(4a)sin(2(ax+b))+C