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

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

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

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

sec (x) - tan (x) + C

Solution steps

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

Expand

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

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

\int \frac{sin ( x )}{cos ^{2} ( x )} d x = sec (x)

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

= sec (x) - tan (x)

Add a constant to the solution

= sec (x) - tan (x) + C

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