3.3 Q-16

Question Statement

Evaluate the following integral:

$$\int \cos x \left( \frac{\ln \sin x}{\sin x} \right) , dx$$

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Background and Explanation

This problem requires the use of substitution and integration by parts. The key is to recognize the logarithmic term $\ln \sin x$ and apply a substitution to simplify the integral. Once the substitution is made, the integral becomes easier to solve.

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Solution

Let’s go through the solution step by step:

1. Rewrite the Integral: The given integral is:

$$\int \cos x \left( \frac{\ln \sin x}{\sin x} \right) , dx$$

We can rewrite the integrand as:

$$\int \ln \sin x \times \left( \frac{\cos x}{\sin x} \right) , dx$$

2. Substitute: Let’s introduce a substitution. Define:

$$u = \ln \sin x$$

To find $du$, differentiate both sides with respect to $x$:

$$du = \frac{1}{\sin x} \cos x , dx$$

Now the integral becomes:

$$\int u , du$$

3. Integrate: The integral of $u$ is straightforward:

$$\frac{u^2}{2} + C$$

4. Substitute Back: Recall that $u = \ln \sin x$. Substituting this back into the result gives:

$$\frac{1}{2} (\ln \sin x)^2 + C$$

So, the final result is:

$$\frac{1}{2} (\ln \sin x)^2 + C$$

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Key Formulas or Methods Used

• Substitution:
  • $u = \ln \sin x$, simplifying the integral into a form we can easily integrate.
• Standard Integration:
  • The integral of $u$ is $\frac{u^2}{2} + C$.

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Summary of Steps

1. Rewrite the integral in terms of $\ln \sin x$.
2. Use the substitution $u = \ln \sin x$ and find $du = \frac{\cos x}{\sin x} , dx$.
3. Integrate $u$ to get $\frac{u^2}{2} + C$.
4. Substitute back $u = \ln \sin x$ to get the final result:

$$\frac{1}{2} (\ln \sin x)^2 + C$$