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

The integral of cos^2(ln(b)) is 1/10 (5b+2bsin(2ln(b))+bcos(2ln(b)))+C

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integral of cos^2(ln(b))

\int cos^{2} (ln (b)) d b

\frac{1}{10} (5 b + 2 b sin (2 ln (b)) + b cos (2 ln (b))) + C

Solution steps

\int cos^{2} (ln (b)) d b

Rewrite using trig identities

= \int \frac{1 + cos ( 2 ln ( b ) )}{2} d b

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 + cos (2 ln (b)) d b

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}{2} (\int 1 d b + \int cos (2 ln (b)) d b)

\int 1 d b = b

\int cos (2 ln (b)) d b = \frac{2}{5} b sin (2 ln (b)) + \frac{1}{5} b cos (2 ln (b))

= \frac{1}{2} (b + \frac{2}{5} b sin (2 ln (b)) + \frac{1}{5} b cos (2 ln (b)))

Simplify

\frac{1}{2} (b + \frac{2}{5} b sin (2 ln (b)) + \frac{1}{5} b cos (2 ln (b))) : \frac{1}{10} (5 b + 2 b sin (2 ln (b)) + b cos (2 ln (b)))

= \frac{1}{10} (5 b + 2 b sin (2 ln (b)) + b cos (2 ln (b)))

Add a constant to the solution

= \frac{1}{10} (5 b + 2 b sin (2 ln (b)) + b cos (2 ln (b))) + C

\int \frac{1}{x ^{2} - 2 x - 3} d x

\int \frac{x}{x ^{2} - 5 x + 3} d x

\int 1 - sin^{3} (x) d x

\int \frac{2 x ^{2}}{x ^{2} + 3} d x

\int \frac{10}{3 x} d x

What is the integral of cos^2(ln(b)) ?
The integral of cos^2(ln(b)) is 1/10 (5b+2bsin(2ln(b))+bcos(2ln(b)))+C