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

The integral of (sec(5x)+csc(5x))^2 is 1/5 (tan(5x)+ln(tan^2(5x))-cot(5x))+C

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integral of (sec(5x)+csc(5x))^2

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

\frac{1}{5} (tan (5 x) + ln (tan^{2} (5 x)) - cot (5 x)) + C

Solution steps

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

Expand

(sec (5 x) + csc (5 x))^{2} : sec^{2} (5 x) + 2 sec (5 x) csc (5 x) + csc^{2} (5 x)

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

Apply u-substitution

: \int (sec^{2} (u) + 2 sec (u) csc (u) + csc^{2} (u)) \frac{1}{5} d u

= \int (sec^{2} (u) + 2 sec (u) csc (u) + csc^{2} (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^{2} (u) + 2 sec (u) csc (u) + csc^{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}{5} (\int sec^{2} (u) d u + \int 2 sec (u) csc (u) d u + \int csc^{2} (u) d u)

\int sec^{2} (u) d u = tan (u)

\int 2 sec (u) csc (u) d u = ln (tan^{2} (u))

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{5} (tan (u) + ln (tan^{2} (u)) - cot (u))

Substitute back

u = 5 x

= \frac{1}{5} (tan (5 x) + ln (tan^{2} (5 x)) - cot (5 x))

Add a constant to the solution

= \frac{1}{5} (tan (5 x) + ln (tan^{2} (5 x)) - cot (5 x)) + C

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n = 8 \sum \infty cos (\frac{1}{n ^{3}})

\frac{x - 2}{3 x ^{2} e ^{x}}

\frac{\partial}{\partial x} (3 x + 4 t)

f (x) = sin (3 ln (x))

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