What is the integral of sin(θ/2)tan(θ/2) ?

The integral of sin(θ/2)tan(θ/2) is 2(ln|tan(θ/2)+sec(θ/2)|-sin(θ/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(θ/2)tan(θ/2)

\int sin (\frac{θ}{2}) tan (\frac{θ}{2}) d θ

2 (ln tan (\frac{θ}{2}) + sec (\frac{θ}{2}) - sin (\frac{θ}{2})) + C

Solution steps

\int sin (\frac{θ}{2}) tan (\frac{θ}{2}) d θ

Apply u-substitution

: \int sin (u) tan (u) \cdot 2 d u

= \int sin (u) tan (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int sin (u) tan (u) d u

Rewrite using trig identities

= 2 \cdot \int \frac{1 - cos ^{2} ( u )}{cos ( u )} d u

Expand

\frac{1 - cos ^{2} ( u )}{cos ( u )} : \frac{1}{cos ( u )} - cos (u)

= 2 \cdot \int \frac{1}{cos ( u )} - cos (u) 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

= 2 (\int \frac{1}{cos ( u )} d u - \int cos (u) d u)

\int \frac{1}{cos ( u )} d u = ln ∣ tan (u) + sec (u) ∣

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

= 2 (ln ∣ tan (u) + sec (u) ∣ - sin (u))

Substitute back

u = \frac{θ}{2}

= 2 (ln tan (\frac{θ}{2}) + sec (\frac{θ}{2}) - sin (\frac{θ}{2}))

Add a constant to the solution

= 2 (ln tan (\frac{θ}{2}) + sec (\frac{θ}{2}) - sin (\frac{θ}{2})) + C

\int (7 - csc (x) cot (x)) d x

\int \frac{sin ( 2 x )}{y} + x d x

\int \frac{2}{e ^{2 x} + 6} + \frac{1}{e ^{2 x} - 3} d x

\int 2 π cos (n x) d x

\int t^{5} cos (t^{6}) d t

What is the integral of sin(θ/2)tan(θ/2) ?
The integral of sin(θ/2)tan(θ/2) is 2(ln|tan(θ/2)+sec(θ/2)|-sin(θ/2))+C