What is the integral of e^{t/2}cos(2t) ?

The integral of e^{t/2}cos(2t) is 2(4/17 e^{t/2}sin(2t)+1/17 e^{t/2}cos(2t))+C

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integral of e^{t/2}cos(2t)

\int e^{\frac{t}{2}} cos (2 t) d t

2 (\frac{4}{17} e^{\frac{t}{2}} sin (2 t) + \frac{1}{17} e^{\frac{t}{2}} cos (2 t)) + C

Solution steps

\int e^{\frac{t}{2}} cos (2 t) d t

Apply u-substitution

: \int \frac{1}{2} e^{\frac{u}{4}} cos (u) d u

= \int \frac{1}{2} e^{\frac{u}{4}} cos (u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{\frac{u}{4}} cos (u) d u

Apply u-substitution

: \int e^{v} cos (4 v) \cdot 4 d v

= \frac{1}{2} \cdot \int e^{v} cos (4 v) \cdot 4 d v

Take the constant out:

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

= \frac{1}{2} \cdot 4 \cdot \int e^{v} cos (4 v) d v

Apply Integration By Parts

: e^{v} \frac{1}{4} sin (4 v) - \int e^{v} \frac{1}{4} sin (4 v) d v

= \frac{1}{2} \cdot 4 (e^{v} \frac{1}{4} sin (4 v) - \int e^{v} \frac{1}{4} sin (4 v) d v)

Take the constant out:

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

= \frac{1}{2} \cdot 4 (e^{v} \frac{1}{4} sin (4 v) - \frac{1}{4} \cdot \int e^{v} sin (4 v) d v)

Apply Integration By Parts

: - \frac{1}{4} e^{v} cos (4 v) - \int - \frac{1}{4} e^{v} cos (4 v) d v

= \frac{1}{2} \cdot 4 (e^{v} \frac{1}{4} sin (4 v) - \frac{1}{4} (- \frac{1}{4} e^{v} cos (4 v) - \int - \frac{1}{4} e^{v} cos (4 v) d v))

Take the constant out:

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

= \frac{1}{2} \cdot 4 (e^{v} \frac{1}{4} sin (4 v) - \frac{1}{4} (- \frac{1}{4} e^{v} cos (4 v) - (- \frac{1}{4} \cdot \int e^{v} cos (4 v) d v)))

Therefore

4 \cdot \frac{1}{2} \cdot \int e^{v} cos (4 v) d v = \frac{1}{2} \cdot 4 (e^{v} \frac{1}{4} sin (4 v) - \frac{1}{4} (- \frac{1}{4} e^{v} cos (4 v) - (- \frac{1}{4} \cdot \int e^{v} cos (4 v) d v)))

Isolate

\int e^{v} cos (4 v) d v

= 4 \cdot \frac{1}{2} (\frac{4 e ^{v} sin ( 4 v )}{17} + \frac{e ^{v} cos ( 4 v )}{17})

Substitute back

= 4 \cdot \frac{1}{2} (\frac{4 e ^{\frac{2 t}{4}} sin ( 4 \cdot \frac{2 t}{4} )}{17} + \frac{e ^{\frac{2 t}{4}} cos ( 4 \cdot \frac{2 t}{4} )}{17})

Simplify

4 \cdot \frac{1}{2} (\frac{4 e ^{\frac{2 t}{4}} sin ( 4 \cdot \frac{2 t}{4} )}{17} + \frac{e ^{\frac{2 t}{4}} cos ( 4 \cdot \frac{2 t}{4} )}{17}) : 2 (\frac{4}{17} e^{\frac{t}{2}} sin (2 t) + \frac{1}{17} e^{\frac{t}{2}} cos (2 t))

= 2 (\frac{4}{17} e^{\frac{t}{2}} sin (2 t) + \frac{1}{17} e^{\frac{t}{2}} cos (2 t))

Add a constant to the solution

= 2 (\frac{4}{17} e^{\frac{t}{2}} sin (2 t) + \frac{1}{17} e^{\frac{t}{2}} cos (2 t)) + C

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

(20.8) (9.16)

\int 20 x^{3} + 12 x^{2} + 6 d x

\int cos^{3} (x) ln (sin (x)) d x

f (x) = \frac{3 x + 2}{3 x - 2}

What is the integral of e^{t/2}cos(2t) ?
The integral of e^{t/2}cos(2t) is 2(4/17 e^{t/2}sin(2t)+1/17 e^{t/2}cos(2t))+C