What is the integral of (sec^3(θ))/(tan(θ)) ?

The integral of (sec^3(θ))/(tan(θ)) is sec(θ)+ln|tan(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 (sec^3(θ))/(tan(θ))

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

sec (θ) + ln tan (\frac{θ}{2}) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{sec ^{3} ( θ ) cos ( θ )}{sin ( θ )} d θ

Use the basic trigonometric identity:

\frac{cos ( x )}{sin ( x )} = cot (x)

= \int sec^{3} (θ) cot (θ) d θ

Rewrite using trig identities

= \int \frac{sec ^{2} ( θ )}{sin ( θ )} d θ

Apply Integration By Parts

: sec (θ) - \int - cot (θ) csc (θ) tan (θ) d θ

= sec (θ) - \int - cot (θ) csc (θ) tan (θ) d θ

\int - cot (θ) csc (θ) tan (θ) d θ = - ln tan (\frac{θ}{2})

= sec (θ) - (- ln tan (\frac{θ}{2}))

Simplify

= sec (θ) + ln tan (\frac{θ}{2})

Add a constant to the solution

= sec (θ) + ln tan (\frac{θ}{2}) + C

f (x) = 16 cos (2 x)

(e^{x} + y) d x + (2 + x + y e^{y}) d y = 0, y (0) = 1

n = 0 \sum \infty n \cdot (n + 1) \cdot x^{n}

\int \frac{1}{16 + 6 x - x ^{2}} d x

\frac{\partial}{\partial x} (\frac{y ^{4}}{x})

What is the integral of (sec^3(θ))/(tan(θ)) ?
The integral of (sec^3(θ))/(tan(θ)) is sec(θ)+ln|tan(θ/2)|+C