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

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

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

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

\frac{1}{2} (tan (x) - sec (x)) + C

Solution steps

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

Take the constant out:

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

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

Expand

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

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

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

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

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

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

Add a constant to the solution

= \frac{1}{2} (tan (x) - sec (x)) + C

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