What is the integral of e^{2x}cos(h)x ?

The integral of e^{2x}cos(h)x is cos(h) 1/4 (e^{2x}*2x-e^{2x})+C

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integral of e^{2x}cos(h)x

\int e^{2 x} cos (h) x d x

cos (h) \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) + C

Solution steps

\int e^{2 x} cos (h) x d x

Take the constant out:

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

= cos (h) \cdot \int e^{2 x} x d x

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

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

Take the constant out:

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

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

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

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

\int e^{u} d u = e^{u}

= cos (h) \frac{1}{4} (e^{u} u - e^{u})

Substitute back

u = 2 x

= cos (h) \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

Add a constant to the solution

= cos (h) \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) + C

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What is the integral of e^{2x}cos(h)x ?
The integral of e^{2x}cos(h)x is cos(h) 1/4 (e^{2x}*2x-e^{2x})+C