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))

integral

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

- (f (x))^{2}

\frac{( x ^{4} )}{( x ^{5} + x )}

y = \frac{4}{( 5 - 3 x )}

x (tan (x))^{2}

x^{2} + 2

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