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

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

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integral of (1/(cos^2(x))-1)cos(x))

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

d (ln ∣ tan (x) + sec (x) ∣ - sin (x)) + C

Solution steps

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

Take the constant out:

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

= d \cdot \int (\frac{1}{cos ^{2} ( x )} - 1) cos (x) d x

Simplify

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

= d \cdot \int sin (x) tan (x) d x

Rewrite using trig identities

= d \cdot \int \frac{1 - cos ^{2} ( x )}{cos ( x )} d x

Expand

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

= d \cdot \int \frac{1}{cos ( x )} - cos (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

= d (\int \frac{1}{cos ( x )} d x - \int cos (x) d x)

\int \frac{1}{cos ( x )} d x = ln ∣ tan (x) + sec (x) ∣

\int cos (x) d x = sin (x)

= d (ln ∣ tan (x) + sec (x) ∣ - sin (x))

Add a constant to the solution

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

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