What is the integral from 1 to 2 of sin(x)+tan(x) ?

The integral from 1 to 2 of sin(x)+tan(x) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of sin(x)+tan(x)

\int_{1}^{2} sin (x) + tan (x) d x

diverges

Solution steps

\int_{1}^{2} sin (x) + tan (x) d x

Take the denominator(s) of

sin (x) + tan (x)

and compare to zero:

x = \frac{π}{2}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{1}^{\frac{π}{2}} sin (x) + tan (x) d x + \int_{\frac{π}{2}}^{2} sin (x) + tan (x) d x

\int_{1}^{\frac{π}{2}} sin (x) + tan (x) d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int (x - 1) (6 x - 5) d x

\frac{d}{d x} (sin^{2} (2 x + 5))

\frac{d}{d x} ((\frac{x - 7}{x + 2})^{3})

What is the integral from 1 to 2 of sin(x)+tan(x) ?
The integral from 1 to 2 of sin(x)+tan(x) is diverges