Question

Integrate the following problem:
$\int {e^{-x} \cdot cos(2x)} \, dx$
P.S:
Have fun!​

Answer 1

We're asked to evaluate the integral,

$\displaystyle\longrightarrow I=\int e^{-x}\cos(2x)\ dx$

As per ILATE rule, it's taken as,

$\displaystyle\longrightarrow I=\int\cos(2x)\cdot e^{-x}\ dx\quad\quad\dots(1)$

Performing integration by parts,

$\displaystyle\longrightarrow I=-e^{-x}\cos(2x)-2\int\sin(2x)\cdot e^{-x}\ dx$

Again performing integration by parts,

$\displaystyle\longrightarrow I=-e^{-x}\cos(2x)-2\left[-e^{-x}\sin(2x)+2\int\cos(2x)\cdot e^{-x}\ dx\right]$

$\displaystyle\longrightarrow I=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)-4\int\cos(2x)\cdot e^{-x}\ dx$

From (1),

$\displaystyle\longrightarrow I=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)-4I$

$\displaystyle\longrightarrow 5I=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{e^{-x}\left(2\sin(2x)-\cos(2x)\right)}{5}(+C)}}$

Answer 2

Answer:

We're asked to evaluate the integrate,

\displaystyle\longrightarrow I=\int e^{-x}\cos(2x)\ dx⟶I=∫e

−x

cos(2x) dx

As per ILATE rule, it's taken as,

\displaystyle\longrightarrow I=\int\cos(2x)\cdot e^{-x}\ dx\quad\quad\dots(1)⟶I=∫cos(2x)⋅e

−x

dx…(1)

Performing integrating by parts,

\displaystyle\longrightarrow I=-e^{-x}\cos(2x)-2\int\sin(2x)\cdot e^{-x}\ dx⟶I=−e

−x

cos(2x)−2∫sin(2x)⋅e

−x

dx

Again performing integrating by parts,

\displaystyle\longrightarrow I=-e^{-x}\cos(2x)-2\left[-e^{-x}\sin(2x)+2\int\cos(2x)\cdot e^{-x}\ dx\right]⟶I=−e

−x

cos(2x)−2[−e

−x

sin(2x)+2∫cos(2x)⋅e

−x

dx]

\displaystyle\longrightarrow I=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)-4\int\cos(2x)\cdot e^{-x}\ dx⟶I=−e

−x

cos(2x)+2e

−x

sin(2x)−4∫cos(2x)⋅e

−x

dx

From (1),

\displaystyle\longrightarrow I=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)-4I⟶I=−e

−x

cos(2x)+2e

−x

sin(2x)−4I

\displaystyle\longrightarrow 5I=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)⟶5I=−e

−x

cos(2x)+2e

−x

sin(2x)

\displaystyle\longrightarrow\underline{\underline{I=\dfrac{e^{-x}\left(2\sin(2x)-\cos(2x)\right)}{5}(+C)}}⟶

I=

5

e

−x

(2sin(2x)−cos(2x))

(+C)