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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= 2 \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

= 2 \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

= 2 \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)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{sin ( 2 x )}{( 1 + cos ( 2 x ) ) ^{2}} d x

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

\int \frac{tan ( x )}{1 - tan ^{2} ( x )} d x

\int 3 x^{2} - 2 x + 1 - \frac{1}{x ^{2}} d x

\int x - 2 ∣ x ∣ d x

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