What is the integral of e^tsqrt(9-e^{2t)} ?

The integral of e^tsqrt(9-e^{2t)} is 9/2 (arcsin(1/3 e^t)+1/2 sin(2arcsin(1/3 e^t)))+C

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integral of e^tsqrt(9-e^{2t)}

\int e^{t} 9 - e^{2 t} d t

\frac{9}{2} (arcsin (\frac{1}{3} e^{t}) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} e^{t}))) + C

Solution steps

\int e^{t} 9 - e^{2 t} d t

Apply u-substitution

: \int 9 - u^{2} d u

= \int 9 - u^{2} d u

Apply Trigonometric Substitution

: \int 9 cos^{2} (v) d v

= \int 9 cos^{2} (v) d v

Take the constant out:

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

= 9 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 9 \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

= 9 \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

= 9 \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)

= 9 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} e^{t}) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} e^{t})))

Simplify

9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} e^{t}) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} e^{t}))) : \frac{9}{2} (arcsin (\frac{1}{3} e^{t}) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} e^{t})))

= \frac{9}{2} (arcsin (\frac{1}{3} e^{t}) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} e^{t})))

Add a constant to the solution

= \frac{9}{2} (arcsin (\frac{1}{3} e^{t}) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} e^{t}))) + C

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What is the integral of e^tsqrt(9-e^{2t)} ?
The integral of e^tsqrt(9-e^{2t)} is 9/2 (arcsin(1/3 e^t)+1/2 sin(2arcsin(1/3 e^t)))+C