What is the integral of e^{2x}sqrt(16-e^{4x)} ?

The integral of e^{2x}sqrt(16-e^{4x)} is 4(arcsin(1/4 e^{2x})+1/2 sin(2arcsin(1/4 e^{2x})))+C

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integral of e^{2x}sqrt(16-e^{4x)}

\int e^{2 x} 16 - e^{4 x} d x

4 (arcsin (\frac{1}{4} e^{2 x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} e^{2 x}))) + C

Solution steps

\int e^{2 x} 16 - e^{4 x} d x

Apply u-substitution

: \int \frac{1}{4} e^{\frac{u}{2}} 16 - e^{u} d u

= \int \frac{1}{4} e^{\frac{u}{2}} 16 - e^{u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{\frac{u}{2}} 16 - e^{u} d u

Apply u-substitution

: \int e^{v} 16 - e^{2 v} \cdot 2 d v

= \frac{1}{4} \cdot \int e^{v} 16 - e^{2 v} \cdot 2 d v

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int e^{v} 16 - e^{2 v} d v

Apply u-substitution

: \int 16 - w^{2} d w

= \frac{1}{4} \cdot 2 \cdot \int 16 - w^{2} d w

Apply Trigonometric Substitution

: \int 16 cos^{2} (t) d t

= \frac{1}{4} \cdot 2 \cdot \int 16 cos^{2} (t) d t

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot 16 \cdot \int cos^{2} (t) d t

Rewrite using trig identities

= \frac{1}{4} \cdot 2 \cdot 16 \cdot \int \frac{1 + cos ( 2 t )}{2} d t

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot 16 \cdot \frac{1}{2} \cdot \int 1 + cos (2 t) d t

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}{4} \cdot 2 \cdot 16 \cdot \frac{1}{2} (\int 1 d t + \int cos (2 t) d t)

\int 1 d t = t

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

= \frac{1}{4} \cdot 2 \cdot 16 \cdot \frac{1}{2} (t + \frac{1}{2} sin (2 t))

Substitute back

= \frac{1}{4} \cdot 2 \cdot 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} e^{\frac{4 x}{2}}) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} e^{\frac{4 x}{2}})))

Simplify

\frac{1}{4} \cdot 2 \cdot 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} e^{\frac{4 x}{2}}) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} e^{\frac{4 x}{2}}))) : 4 (arcsin (\frac{1}{4} e^{2 x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} e^{2 x})))

= 4 (arcsin (\frac{1}{4} e^{2 x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} e^{2 x})))

Add a constant to the solution

= 4 (arcsin (\frac{1}{4} e^{2 x}) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} e^{2 x}))) + C

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What is the integral of e^{2x}sqrt(16-e^{4x)} ?
The integral of e^{2x}sqrt(16-e^{4x)} is 4(arcsin(1/4 e^{2x})+1/2 sin(2arcsin(1/4 e^{2x})))+C