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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int x^{4} + 7 x d x

\int \frac{sec ^{3} ( x )}{tan ^{3} ( x )} d x

\int \frac{1}{3 x + 3} d x

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

\int (x - 1) (x - 2) d x

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