What is the integral of arccsc(nt) ?

The integral of arccsc(nt) is tarccsc(nt)+1/n ln|nt+sqrt(n^2t^2-1)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of arccsc(nt)

\int arccsc (n t) d t

t arccsc (n t) + \frac{1}{n} ln n t + n^{2} t^{2} - 1 + C

Solution steps

\int arccsc (n t) d t

Apply u-substitution

: \int \frac{arccsc ( u )}{n} d u

= \int \frac{arccsc ( u )}{n} d u

Take the constant out:

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

= \frac{1}{n} \cdot \int arccsc (u) d u

Apply Integration By Parts

: u arccsc (u) - \int - \frac{u}{u ^{2} u ^{2} - 1} d u

= \frac{1}{n} (u arccsc (u) - \int - \frac{u}{u ^{2} u ^{2} - 1} d u)

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

= \frac{1}{n} (u arccsc (u) - (- ln u + u^{2} - 1))

Substitute back

u = n t

= \frac{1}{n} (n t arccsc (n t) - (- ln n t + (n t)^{2} - 1))

Simplify

\frac{1}{n} (n t arccsc (n t) - (- ln n t + (n t)^{2} - 1)) : t arccsc (n t) + \frac{1}{n} ln n t + n^{2} t^{2} - 1

= t arccsc (n t) + \frac{1}{n} ln n t + n^{2} t^{2} - 1

Add a constant to the solution

= t arccsc (n t) + \frac{1}{n} ln n t + n^{2} t^{2} - 1 + C

\int \frac{( 4 - x )}{2 - x ^{2}} d x

\int 576 - 36 x^{2} d x

\int x^{2} (x^{3} - 3)^{4} d x

\int (x^{2} + 3 x - 2) e^{2 x} d x

\int (4 - x^{2}) (2 x - x^{3}) d x

What is the integral of arccsc(nt) ?
The integral of arccsc(nt) is tarccsc(nt)+1/n ln|nt+sqrt(n^2t^2-1)|+C