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

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

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

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

2 tan (x) - 2 x + C

Solution steps

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

Expand

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

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

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

= tan (x) - (- tan (x) + 2 x)

Simplify

= 2 tan (x) - 2 x

Add a constant to the solution

= 2 tan (x) - 2 x + C

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