What is the integral of e^xsqrt(81-e^{2x)} ?

The integral of e^xsqrt(81-e^{2x)} is 81/2 (arcsin(1/9 e^x)+1/2 sin(2arcsin(1/9 e^x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^xsqrt(81-e^{2x)}

\int e^{x} 81 - e^{2 x} d x

\frac{81}{2} (arcsin (\frac{1}{9} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{9} e^{x}))) + C

Solution steps

\int e^{x} 81 - e^{2 x} d x

Apply u-substitution

: \int 81 - u^{2} d u

= \int 81 - u^{2} d u

Apply Trigonometric Substitution

: \int 81 cos^{2} (v) d v

= \int 81 cos^{2} (v) d v

Take the constant out:

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

= 81 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 81 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 81 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 81 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 81 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 81 \cdot \frac{1}{2} (arcsin (\frac{1}{9} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{9} e^{x})))

Simplify

81 \cdot \frac{1}{2} (arcsin (\frac{1}{9} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{9} e^{x}))) : \frac{81}{2} (arcsin (\frac{1}{9} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{9} e^{x})))

= \frac{81}{2} (arcsin (\frac{1}{9} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{9} e^{x})))

Add a constant to the solution

= \frac{81}{2} (arcsin (\frac{1}{9} e^{x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{9} e^{x}))) + C

y^{^{'}} - y = 3 t e^{2 t}

\frac{\partial}{\partial x} (3 x + 2 y)

\int \frac{1}{9 x ^{2} + 6 x - 8} d x

n = 0 \sum \infty n \frac{1}{5 ^{n}}

\int sec^{2} (3 x t a) n^{5} 3 x d x

What is the integral of e^xsqrt(81-e^{2x)} ?
The integral of e^xsqrt(81-e^{2x)} is 81/2 (arcsin(1/9 e^x)+1/2 sin(2arcsin(1/9 e^x)))+C