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

The integral of 1/((1+cos(x))^2) is 1/2 (tan(x/2)+(tan^3(x/2))/3)+C

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

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

\frac{1}{2} (tan (\frac{x}{2}) + \frac{tan ^{3} ( \frac{x}{2} )}{3}) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

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

\int 1 d u = u

\int u^{2} d u = \frac{u ^{3}}{3}

= \frac{1}{2} (u + \frac{u ^{3}}{3})

Substitute back

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

= \frac{1}{2} (tan (\frac{x}{2}) + \frac{tan ^{3} ( \frac{x}{2} )}{3})

Add a constant to the solution

= \frac{1}{2} (tan (\frac{x}{2}) + \frac{tan ^{3} ( \frac{x}{2} )}{3}) + C

\int \frac{1}{t + 2} d t

\int_{- 1}^{2} x^{2} e^{x^{3} + 1} d x

f (x) = ln (sec (x))

\int_{1}^{x} t^{4} d t

f (x) = (x^{2} - 2)^{2}

What is the integral of 1/((1+cos(x))^2) ?
The integral of 1/((1+cos(x))^2) is 1/2 (tan(x/2)+(tan^3(x/2))/3)+C