Salinas Bush
12/24/2023 · Senior High School
8:- Attempt any one:- (i) Evaluate \( \int e^{x} \sin x d x \)
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Tutor-Verified Answer
Step-by-step Solution
To evaluate the integral \( \int e^{x} \sin x \, dx \), we can use the method of integration by parts or the technique of integration by parts twice. Here, we will use integration by parts.
Let:
- \( u = \sin x \) and \( dv = e^x \, dx \)
Then, we differentiate and integrate:
- \( du = \cos x \, dx \)
- \( v = e^x \)
Now, applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[
\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx
\]
Next, we need to evaluate \( \int e^x \cos x \, dx \). We will again use integration by parts. Let:
- \( u = \cos x \) and \( dv = e^x \, dx \)
Then:
- \( du = -\sin x \, dx \)
- \( v = e^x \)
Applying integration by parts again:
\[
\int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx
\]
\[
= e^x \cos x + \int e^x \sin x \, dx
\]
Now we have two equations:
1. \( \int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx \)
2. \( \int e^x \cos x \, dx = e^x \cos x + \int e^x \sin x \, dx \)
Substituting the second equation into the first:
\[
\int e^x \sin x \, dx = e^x \sin x - \left( e^x \cos x + \int e^x \sin x \, dx \right)
\]
This simplifies to:
\[
\int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx
\]
Now, we can add \( \int e^x \sin x \, dx \) to both sides:
\[
2 \int e^x \sin x \, dx = e^x \sin x - e^x \cos x
\]
Dividing both sides by 2:
\[
\int e^x \sin x \, dx = \frac{1}{2} (e^x \sin x - e^x \cos x)
\]
Finally, we can add the constant of integration \( C \):
\[
\int e^x \sin x \, dx = \frac{1}{2} e^x (\sin x - \cos x) + C
\]
Thus, the evaluated integral is:
\[
\int e^x \sin x \, dx = \frac{1}{2} e^x (\sin x - \cos x) + C
\]
Quick Answer
The integral \( \int e^{x} \sin x \, dx \) evaluates to \( \frac{1}{2} e^x (\sin x - \cos x) + C \).
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