What is the integral of-(sec^2(x)-1)/(tan(x)) ?

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

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integral of-(sec^2(x)-1)/(tan(x))

\int - \frac{sec ^{2} ( x ) - 1}{tan ( x )} d x

- ln ∣ tan (x) ∣ + ln ∣ sin (x) ∣ + C

Solution steps

\int - \frac{sec ^{2} ( x ) - 1}{tan ( x )} d x

Take the constant out:

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

= - \int \frac{sec ^{2} ( x ) - 1}{tan ( x )} d x

Expand

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

= - \int \frac{sec ^{2} ( x )}{tan ( x )} - \frac{1}{tan ( x )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int \frac{sec ^{2} ( x )}{tan ( x )} d x - \int \frac{1}{tan ( x )} d x)

\int \frac{sec ^{2} ( x )}{tan ( x )} d x = ln ∣ tan (x) ∣

\int \frac{1}{tan ( x )} d x = ln ∣ sin (x) ∣

= - (ln ∣ tan (x) ∣ - ln ∣ sin (x) ∣)

Simplify

= - ln ∣ tan (x) ∣ + ln ∣ sin (x) ∣

Add a constant to the solution

= - ln ∣ tan (x) ∣ + ln ∣ sin (x) ∣ + C

\frac{s + 3}{s ^{2} + s + 1}

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{5}{x ^{4}}) (x + 9 x^{3}))

x \to 0 + lim (\frac{( 3 )}{x ^{2} - x})

\int_{0}^{8} \frac{x}{( x + 8 ) ^{3}} d x

\frac{d}{d x} ((3 x^{2} - 5 x) e^{x})

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