What is the integral of cos(5x)sec^2(5x) ?

The integral of cos(5x)sec^2(5x) is 1/5 ln|tan(5x)+sec(5x)|+C

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integral of cos(5x)sec^2(5x)

\int cos (5 x) sec^{2} (5 x) d x

\frac{1}{5} ln ∣ tan (5 x) + sec (5 x) ∣ + C

Solution steps

\int cos (5 x) sec^{2} (5 x) d x

Rewrite using trig identities

= \int sec (5 x) d x

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = 5 x

= \frac{1}{5} ln ∣ tan (5 x) + sec (5 x) ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ tan (5 x) + sec (5 x) ∣ + C

(3 x^{2} + 9) (x^{2} - 1 x)

(2 x y - 9 x^{2}) + (2 y + x^{2} + 1) \frac{d y}{d x} = 0

f (x) = - 8 x^{2} + 8

\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{1}{3}} + 5

y = x - 1, y = 0, x = 9

What is the integral of cos(5x)sec^2(5x) ?
The integral of cos(5x)sec^2(5x) is 1/5 ln|tan(5x)+sec(5x)|+C