What is the integral of 1/(3cos(5x-pi/4)) ?

The integral of 1/(3cos(5x-pi/4)) is 1/15 ln|tan(5x-pi/4)+sec(5x-pi/4)|+C

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integral of 1/(3cos(5x-pi/4))

\int \frac{1}{3 cos ( 5 x - \frac{π}{4} )} d x

\frac{1}{15} ln tan (5 x - \frac{π}{4}) + sec (5 x - \frac{π}{4}) + C

Solution steps

\int \frac{1}{3 cos ( 5 x - \frac{π}{4} )} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{cos ( 5 x - \frac{π}{4} )} d x

Rewrite using trig identities

= \frac{1}{3} \cdot \int sec (5 x - \frac{π}{4}) d x

Apply u-substitution

: \int sec (u) \frac{1}{5} d u

= \frac{1}{3} \cdot \int sec (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{3} \cdot \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = 5 x - \frac{π}{4}

= \frac{1}{3} \cdot \frac{1}{5} ln tan (5 x - \frac{π}{4}) + sec (5 x - \frac{π}{4})

Simplify

\frac{1}{3} \cdot \frac{1}{5} ln tan (5 x - \frac{π}{4}) + sec (5 x - \frac{π}{4}) : \frac{1}{15} ln tan (5 x - \frac{π}{4}) + sec (5 x - \frac{π}{4})

= \frac{1}{15} ln tan (5 x - \frac{π}{4}) + sec (5 x - \frac{π}{4})

Add a constant to the solution

= \frac{1}{15} ln tan (5 x - \frac{π}{4}) + sec (5 x - \frac{π}{4}) + C

\frac{d}{d x} ((x)^{l n (x)})

\int arctan (y) d y

f (x) = x^{4} + 5 e^{x}, (0, 5)

cos^{2} (x) sin (x) (\frac{d y}{d x}) + (cos^{3} (x)) y = 1

x \to 0 lim (x^{n} \cdot cos (\frac{3}{x}))

What is the integral of 1/(3cos(5x-pi/4)) ?
The integral of 1/(3cos(5x-pi/4)) is 1/15 ln|tan(5x-pi/4)+sec(5x-pi/4)|+C