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

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

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

ln (sec^{2} (\frac{x}{2})) + tan (\frac{x}{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{1 + u ^{2} + 2 u}{1 + u ^{2}} d u

= \int \frac{1 + u ^{2} + 2 u}{1 + u ^{2}} d u

Expand

\frac{1 + u ^{2} + 2 u}{1 + u ^{2}} : \frac{1}{1 + u ^{2}} + \frac{u ^{2}}{1 + u ^{2}} + \frac{2 u}{1 + u ^{2}}

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 + u ^{2}} d u + \int \frac{u ^{2}}{1 + u ^{2}} d u + \int \frac{2 u}{1 + u ^{2}} d u

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

\int \frac{u ^{2}}{1 + u ^{2}} d u = - arctan (u) + u

\int \frac{2 u}{1 + u ^{2}} d u = ln 1 + u^{2}

= arctan (u) - arctan (u) + u + ln 1 + u^{2}

Substitute back

u = tan (\frac{x}{2})

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

Simplify

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

= ln (sec^{2} (\frac{x}{2})) + tan (\frac{x}{2})

Add a constant to the solution

= ln (sec^{2} (\frac{x}{2})) + tan (\frac{x}{2}) + C

\frac{\partial}{\partial x} (\frac{( y - 3 x )}{z})

f (x) = x^{2} + 8, (- 3, 17)

\int_{2}^{7} \frac{1}{x ^{2} - 1} d x

y = \frac{6 x}{x + 2}

\frac{d}{d x} ((1 + 5 x) (1 - 7 x))

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