3.3 Q-15

Question Statement

Evaluate the following integral:

$$\int \frac{\cos x}{\sin x \ln \sin x} , dx$$

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Background and Explanation

This integral involves the use of substitution and logarithmic properties. The key to solving this is recognizing the structure of the integral, which suggests using a substitution for the logarithmic term $\ln \sin x$. By doing so, we can simplify the integral into a more manageable form.

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Solution

Let’s break down the solution step by step:

1. Simplify the Integral: The given integral is:

$$\int \frac{\cos x}{\sin x \ln \sin x} , dx$$

We can rewrite the integrand as:

$$\int \frac{1}{\ln \sin x} \cdot \frac{\cos x}{\sin x} , dx$$

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

$$u = \ln \sin x$$

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

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

Now the integral becomes:

$$\int \frac{1}{u} , du$$

3. Integrate: The integral of $\frac{1}{u}$ is a standard logarithmic integral:

$$\ln |u| + C$$

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

$$\ln |\ln \sin x| + C$$

So, the final result is:

$$\ln |\ln \sin x| + C$$

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Key Formulas or Methods Used

• Substitution:
  • $u = \ln \sin x$, which simplifies the integral into a form we can easily integrate.
• Logarithmic Integration:
  • The integral of $\frac{1}{u}$ is $\ln |u| + C$.

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Summary of Steps

1. Simplify the integral into a form involving $\frac{1}{\ln \sin x}$.
2. Use the substitution $u = \ln \sin x$ and find $du = \frac{\cos x}{\sin x} , dx$.
3. Integrate $\frac{1}{u}$ to get $\ln |u| + C$.
4. Substitute back $u = \ln \sin x$ to get the final result:

$$\ln |\ln \sin x| + C$$