3.4 Q-3

Question Statement

Show that the following integral is equal to the given result:

$$\int \mathrm{e}^{\mathrm{ax}} \sin(bx) , dx = \frac{\mathrm{e}^{\mathrm{ax}}}{\sqrt{a^{2} + b^{2}}} \sin\left(bx - \tan^{-1}\frac{b}{a}\right) + C$$

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Background and Explanation

To solve this integral, we use integration by parts, a technique based on the formula:

$$\int u , dv = uv - \int v , du$$

We will need to apply integration by parts twice to handle the terms efficiently. The process involves breaking down the integral into simpler parts and using trigonometric identities along with exponential functions.

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Solution

We begin with the integral:

$$I = \int e^{ax} \sin(bx) , dx$$

Step 1: Apply Integration by Parts

We choose the following parts for integration by parts:

• Let $u = \sin(bx)$, and $dv = e^{ax} dx$.

Now, compute $du$ and $v$:

• $du = b \cos(bx) , dx$
• $v = \frac{e^{ax}}{a}$

Using the integration by parts formula:

$$I = \frac{-\cos(bx)}{b} e^{ax} - \int \frac{-\cos(bx)}{b} e^{ax} \cdot a , dx$$

This simplifies to:

$$I = \frac{-\cos(bx)}{b} e^{ax} + \frac{a}{b} \int e^{ax} \cos(bx) , dx$$

Step 2: Apply Integration by Parts Again

Next, we apply integration by parts to the remaining integral $\int e^{ax} \cos(bx) , dx$. Again, let:

• $u = \cos(bx)$, and $dv = e^{ax} dx$.

Now, compute $du$ and $v$:

• $du = -b \sin(bx) , dx$
• $v = \frac{e^{ax}}{a}$

Substitute into the integration by parts formula:

$$\int e^{ax} \cos(bx) , dx = \frac{e^{ax}}{a} \cos(bx) - \int \frac{e^{ax}}{a} (-b \sin(bx)) , dx$$

Simplifying this:

$$\int e^{ax} \cos(bx) , dx = \frac{e^{ax}}{a} \cos(bx) + \frac{b}{a} \int e^{ax} \sin(bx) , dx$$

Step 3: Substitute the New Integral Back

Now, substitute this result back into our original expression for $I$:

$$I = \frac{-\cos(bx)}{b} e^{ax} + \frac{a}{b} \left( \frac{e^{ax}}{a} \cos(bx) + \frac{b}{a} \int e^{ax} \sin(bx) , dx \right)$$

Simplify:

$$I = \frac{-\cos(bx)}{b} e^{ax} + \frac{e^{ax}}{b} \cos(bx) + \frac{a}{b^2} \int e^{ax} \sin(bx) , dx$$

Step 4: Solve for the Integral

Rearrange the equation to isolate the integral:

$$I - \frac{a}{b^2} I = \frac{e^{ax}}{b} \left( a \sin(bx) - b \cos(bx) \right)$$

Factor out $I$:

$$\left(1 + \frac{a^2}{b^2}\right) I = \frac{e^{ax}}{b^2} \left( a \sin(bx) - b \cos(bx) \right)$$

Simplify the expression for $I$:

$$I = \frac{b^2}{a^2 + b^2} \cdot \frac{1}{b^2} \left( a \sin(bx) - b \cos(bx) \right) e^{ax} + C$$

Thus, we have:

$$I = \frac{1}{a^2 + b^2} \left[ a \sin(bx) - b \cos(bx) \right] e^{ax} + C$$

Finally, recognizing the trigonometric identity:

$$\sin \left( bx - \tan^{-1} \frac{b}{a} \right) = \frac{a \sin(bx) - b \cos(bx)}{\sqrt{a^2 + b^2}}$$

We get the final result:

$$I = \frac{e^{ax}}{\sqrt{a^2 + b^2}} \sin \left( bx - \tan^{-1} \frac{b}{a} \right) + C$$

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Key Formulas or Methods Used

• Integration by Parts: $\int u , dv = uv - \int v , du$

• Trigonometric Identity: $\sin \left( bx - \tan^{-1} \frac{b}{a} \right) = \frac{a \sin(bx) - b \cos(bx)}{\sqrt{a^2 + b^2}}$

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Summary of Steps

1. Apply Integration by Parts: Break the integral into two parts, one involving $\cos(bx)$.
2. Apply Integration by Parts Again: Solve the integral of $e^{ax} \cos(bx)$.
3. Combine the Results: Substitute the results back into the original equation.
4. Solve for $I$: Isolate the integral and simplify.
5. Final Simplification: Use the trigonometric identity to express the result in the desired form.

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Reference

By Great Science Academy: