What is the integral of e^xsqrt(1-e^{2x)} ?

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

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integral of e^xsqrt(1-e^{2x)}

\int e^{x} 1 - e^{2 x} d x

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

Solution steps

\int e^{x} 1 - e^{2 x} d x

Apply u-substitution

: \int 1 - u^{2} d u

= \int 1 - u^{2} d u

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Substitute back

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

Add a constant to the solution

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

e^{1 - x}

x \frac{d y}{d x} + y + y^{2} = 0

\frac{( 1 + tan ( x ) ) ^{2}}{sec ( x )}

\int_{1}^{2} (4 x - \frac{2}{x ^{2}}) d x

\int sec^{7} (4 x) tan (4 x) d x

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