Let's evaluate the integral $I = \int 9 \sin(x) \cos(x) \, dx$ using four different methods, as requested:

Method 1: Substitution with $u = \cos(x)$

1. Substitute $u = \cos(x)$, hence $du = -\sin(x) dx$.

   The integral becomes: $I = \int 9 \sin(x) \cos(x) \, dx = -9 \int u \, du$

2. Integrate: $-9 \int u \, du = -9 \left( \frac{u^2}{2} \right) + C = -\frac{9}{2} \cos^2(x) + C$

   Answer 1: $I = -\frac{9}{2} \cos^2(x) + C$

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Method 2: Substitution with $u = \sin(x)$

1. Substitute $u = \sin(x)$, hence $du = \cos(x) dx$.

   The integral becomes: $I = \int 9 u \, du$

2. Integrate: $9 \int u \, du = 9 \left( \frac{u^2}{2} \right) + C = \frac{9}{2} \sin^2(x) + C$

   Answer 2: $I = \frac{9}{2} \sin^2(x) + C$

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Method 3: Using the identity $\sin(2x) = 2 \sin(x) \cos(x)$

We recognize that $\sin(2x) = 2 \sin(x) \cos(x)$, so: $9 \sin(x) \cos(x) = \frac{9}{2} \sin(2x)$

Now, integrate: $I = \int \frac{9}{2} \sin(2x) \, dx$

The integral of $\sin(2x)$ is $-\frac{1}{2} \cos(2x)$, so: $I = \frac{9}{2} \left( -\frac{1}{2} \cos(2x) \right) + C = -\frac{9}{4} \cos(2x) + C$

Answer 3: $I = -\frac{9}{4} \cos(2x) + C$

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Method 4: Integration by Parts

We can apply integration by parts where we let $u = \sin(x)$ and $dv = 9 \cos(x) \, dx$. Then, $du = \cos(x) \, dx$ and $v = 9 \sin(x)$.

Using the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get:

1. $u = \sin(x)$, $dv = 9 \cos(x) \, dx$
2. $v = 9 \sin(x)$, $du = \cos(x) \, dx$

Now: $I = 9 \sin(x) \sin(x) - \int 9 \sin(x) \cos(x) \, dx$

This brings us back to the original integral on the right-hand side. So we use algebra to solve: $I = \frac{9}{2} \sin^2(x) + C$

Answer 4: $I = \frac{9}{2} \sin^2(x) + C$

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Summary of Answers:

1. $I = -\frac{9}{2} \cos^2(x) + C$
2. $I = \frac{9}{2} \sin^2(x) + C$
3. $I = -\frac{9}{4} \cos(2x) + C$
4. $I = \frac{9}{2} \sin^2(x) + C$

Would you like further details or clarification on any of these methods?

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5 Related Questions:

1. How do you choose an appropriate substitution for integration?
2. What is the integration by parts formula, and how is it applied?
3. Can we generalize the identity $\sin(2x) = 2 \sin(x) \cos(x)$ for higher powers of sine and cosine?
4. How does the symmetry of sine and cosine functions affect integration?
5. What is the geometric interpretation of $\sin(2x)$ in integrals?

Tip: Substituting trigonometric identities can often simplify complex integrals and make them easier to compute!