What is the integral of sec(x)csc(x) ?

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

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integral of sec(x)csc(x)

\int sec (x) csc (x) d x

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

Solution steps

\int sec (x) csc (x) d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 1} d u

= \int \frac{u}{u ^{2} - 1} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

= \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln sec^{2} (x) - 1

Simplify

\frac{1}{2} ln sec^{2} (x) - 1 : \frac{1}{2} ln (tan^{2} (x))

= \frac{1}{2} ln (tan^{2} (x))

Add a constant to the solution

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

\int x^{2} - 4 x d x

f (x) = 2 x^{2} - 9 x + 10, at x = 1

y^{^{'}} + \frac{1}{x} y = \frac{2}{3} x^{4} y^{4}

y = x^{2} + 2 x + 11

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