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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int \frac{1}{sec ^{2} ( x )} d x = \frac{1}{2} (x + \frac{1}{2} sin (2 x))

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

Add a constant to the solution

= \frac{sin ^{2} ( x )}{2} - \frac{1}{2} (x + \frac{1}{2} sin (2 x)) + C

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