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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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