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

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

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integral of (1+cos(e^{-2x}))/(e^{2x)}

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - 1 - cos (v) d v

= \frac{1}{2} \cdot \int - 1 - cos (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

= \frac{1}{2} (- \int 1 d v - \int cos (v) d v)

\int 1 d v = v

\int cos (v) d v = sin (v)

= \frac{1}{2} (- v - sin (v))

Substitute back

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

Add a constant to the solution

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

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