Integrals Involving sin x, cos x and Exponential Functions

Let \( u = \sin(x) \) and \( \dfrac{dv}{dx} = e^x \), which gives \( u' = \cos(x) \) and \( v = \displaystyle \int e^x \, dx = e^x \).

Using integration by parts:

\[ \int \sin(x)e^x \, dx = \sin(x)e^x - \int \cos(x)e^x \, dx \]

Apply integration by parts a second time to the term \( \displaystyle \int \cos(x)e^x \, dx \):

\[ \int \sin(x)e^x \, dx = \sin(x)e^x - \left( \cos(x)e^x - \int (-\sin(x))e^x \, dx \right) \] \[ \int \sin(x)e^x \, dx = \sin(x)e^x - \cos(x)e^x - \int \sin(x)e^x \, dx \]

Since the integral on the right is the same as the original integral, we can combine them by adding \( \int \sin(x)e^x \, dx \) to both sides:

\[ 2 \int \sin(x)e^x \, dx = \sin(x)e^x - \cos(x)e^x \]

Dividing by 2 gives the final result:

\[ \int \sin(x)e^x \, dx = \dfrac{1}{2} e^x (\sin(x) - \cos(x)) + C \]

Let \( u = \cos(2x) \) and \( \dfrac{dv}{dx} = e^x \), which gives \( u' = -2\sin(2x) \) and \( v = e^x \).

Apply integration by parts:

\[ \begin{aligned} \int \cos(2x)e^x \, dx &= \cos(2x)e^x - \int e^x (-2\sin(2x)) \, dx \\[6pt] &= \cos(2x)e^x + 2 \int e^x \sin(2x) \, dx \end{aligned} \]

Apply integration by parts to the new integral \( \int e^x \sin(2x) \, dx \) (letting \( w = \sin(2x) \) and \( dw/dx = e^x \)):

\[ \begin{aligned} \int \cos(2x)e^x \, dx &= \cos(2x)e^x + 2 \left( \sin(2x)e^x - 2 \int \cos(2x)e^x \, dx \right) \\[6pt] &= \cos(2x)e^x + 2\sin(2x)e^x - 4 \int \cos(2x)e^x \, dx \end{aligned} \]

Grouping the identical integrals on the left-hand side:

\[ 5 \int \cos(2x)e^x \, dx = e^x (\cos(2x) + 2\sin(2x)) \]

Thus, the integral is:

\[ \int \cos(2x) \, e^x \, dx = \dfrac{1}{5} e^x ( \cos(2x) + 2 \sin(2x) ) + C \]

Let \( u = \sin(3x + 2) \) and \( \dfrac{dv}{dx} = e^{3x} \), which gives \( u' = 3\cos(3x + 2) \) and \( v = \dfrac{1}{3}e^{3x} \).

Apply integration by parts:

\[ \int \sin(3x + 2) \, e^{3x} \, dx = \dfrac{1}{3}\sin(3x + 2)e^{3x} - \int \cos(3x + 2) \left(\dfrac{1}{3}e^{3x}\right) \, dx \]

Apply integration by parts one more time to \( \int \cos(3x + 2) e^{3x} \, dx \):

\[ \int \sin(3x + 2) e^{3x} \, dx = \dfrac{1}{3} \sin(3x + 2) e^{3x} - \left( \dfrac{1}{3}\cos(3x + 2)e^{3x} + \int \sin(3x + 2)e^{3x} \, dx \right) \]

Grouping the integrals:

\[ 2 \int \sin(3x + 2) e^{3x} \, dx = \dfrac{1}{3} e^{3x} \left( \sin(3x + 2) - \cos(3x + 2) \right) \]

Dividing by 2:

\[ \int \sin(3x + 2) e^{3x} \, dx = \dfrac{1}{6} e^{3x} ( \sin(3x + 2) - \cos(3x + 2) ) + C \]

Let \( u = \cos(4x) \) and \( \dfrac{dv}{dx} = e^{2x + 5} \). Applying integration by parts twice:

\[ \begin{aligned} \int \cos(4x)e^{2x + 5} \, dx &= \tfrac{1}{2} e^{2x + 5} \cos(4x) + 2 \int e^{2x + 5} \sin(4x) \, dx \\[6pt] &= \tfrac{1}{2} e^{2x + 5} \cos(4x) + 2 \left( \tfrac{1}{2} e^{2x + 5} \sin(4x) - 2 \int e^{2x + 5} \cos(4x) \, dx \right) \end{aligned} \]

Simplifying and gathering terms:

\[ 5 \int \cos(4x) e^{2x+5} \, dx = \dfrac{1}{2} e^{2x+5} \cos(4x) + e^{2x+5} \sin(4x) \]

Dividing by 5 gives the final solution:

\[ \int \cos(4x) \, e^{2x + 5} \, dx = \dfrac{1}{10} e^{2x + 5} ( \cos(4x) + 2 \sin(4x) ) + C \]