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

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

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

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

\frac{1}{2} (x + sin (x)) + C

Solution steps

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

Rewrite using trig identities

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

Simplify

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

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

Rewrite using trig identities

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

\int 1 d x = x

\int cos (x) d x = sin (x)

= \frac{1}{2} (x + sin (x))

Add a constant to the solution

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

\frac{d}{d x} (10 sin^{2} (x))

\int sec (\frac{π}{4} - 2 x) d x

x \to 3 + lim (x - 2)

f (x) = 5 x^{2}, at x = 10

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

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