Can i please know the answer for this question ???Integral e^ax sin(bx+c) dx

To solve the integral of the form ∫ e^(ax) sin(bx + c) dx, we can use the technique of integration by parts or a more advanced method called the method of undetermined coefficients. This integral is often encountered in engineering and physics, particularly in contexts involving oscillatory motion or wave functions. Let’s break it down step-by-step.

Understanding the Integral

The integral we want to evaluate is:

∫ e^(ax) sin(bx + c) dx

Here, e^(ax) is an exponential function and sin(bx + c) is a sine function, suggesting that the resulting integral may involve a combination of exponential and trigonometric functions.

Applying Integration by Parts

Integration by parts is based on the formula:

∫ u dv = uv - ∫ v du

In our case, we can let:

• u = sin(bx + c) (so that du = b cos(bx + c) dx)
• dv = e^(ax) dx (thus v = (1/a)e^(ax))

Applying this formula, we can start our integration:

∫ e^(ax) sin(bx + c) dx = (1/a)e^(ax) sin(bx + c) - ∫ (1/a)e^(ax) (b cos(bx + c)) dx

Rearranging the Integral

This gives us:

∫ e^(ax) sin(bx + c) dx = (1/a)e^(ax) sin(bx + c) - (b/a) ∫ e^(ax) cos(bx + c) dx

Now, we have a new integral to evaluate:

∫ e^(ax) cos(bx + c) dx

We can apply integration by parts again in a similar manner:

• Let u = cos(bx + c) (then du = -b sin(bx + c) dx)
• Let dv = e^(ax) dx (then v = (1/a)e^(ax))

This leads to:

∫ e^(ax) cos(bx + c) dx = (1/a)e^(ax) cos(bx + c) + (b/a) ∫ e^(ax) sin(bx + c) dx

Combining the Results

We now have two equations to work with:

I = ∫ e^(ax) sin(bx + c) dxJ = ∫ e^(ax) cos(bx + c) dx

Substituting J back into the equation for I gives:

I = (1/a)e^(ax) sin(bx + c) - (b/a) * [(1/a)e^(ax) cos(bx + c) + (b/a) I]

Now we can rearrange this equation to solve for I:

I + (b^2/a^2) I = (1/a)e^(ax) sin(bx + c) - (b/a^2)e^(ax) cos(bx + c)

Factoring I out, we get:

I(1 + (b^2/a^2)) = (1/a)e^(ax) sin(bx + c) - (b/a^2)e^(ax) cos(bx + c)

Finally:

I = [ (1/a)e^(ax) sin(bx + c) - (b/a^2)e^(ax) cos(bx + c)] / [1 + (b^2/a^2)]

Final Result

Thus, the integral of e^(ax) sin(bx + c) is:

∫ e^(ax) sin(bx + c) dx = [ (1/a)e^(ax) sin(bx + c) - (b/a^2)e^(ax) cos(bx + c)] / [1 + (b^2/a^2)] + C

where C is the constant of integration. This formula combines both exponential and trigonometric components, reflecting the nature of the functions involved. You can apply similar steps if you encounter related integrals, adapting the method as necessary.