3.3 Q-12

Question Statement

Evaluate the following integral:

$$\int \frac{\sin \theta , d\theta}{1 + \cos^2 \theta}$$

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Background and Explanation

This integral involves trigonometric functions, and we will simplify it using a substitution. The key concept here is to recognize the trigonometric identity and use substitution to convert the expression into a more familiar form that can be easily integrated.

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Solution

Let’s solve this step by step:

1. Substitute for $x$: We use the substitution $x = \sin \theta$, which implies that $dx = \sin \theta , d\theta$. Now, we can rewrite the integral in terms of $x$.

   The integral becomes:

$$= - \int \frac{1}{1 + \cos^2 \theta} (-\sin \theta) , d\theta$$

2. Simplify the expression: Since $x = \sin \theta$, we can replace $\sin \theta , d\theta$ with $dx$ in the equation. Also, recall that $\cos^2 \theta = 1 - x^2$, so the integral is now:

$$= - \int \frac{1}{1 + x^2} , dx$$

3. Apply the standard integral formula: The integral of $\frac{1}{1 + x^2}$ is a standard result, which gives us the inverse tangent function. Therefore:

$$= - \tan^{-1}(x) + C$$

4. Substitute back $x = \sin \theta$: Finally, substitute back $x = \sin \theta$ to express the solution in terms of $\theta$:

$$= - \tan^{-1}(\sin \theta) + C$$

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Key Formulas or Methods Used

• Substitution:
  • $x = \sin \theta$, $dx = \sin \theta , d\theta$
• Standard Integral:
  • $\int \frac{1}{1 + x^2} , dx = \tan^{-1}(x)$

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Summary of Steps

1. Substitute $x = \sin \theta$, and use $dx = \sin \theta , d\theta$.
2. Rewrite the integral in terms of $x$, simplifying $1 + \cos^2 \theta$ as $1 + x^2$.
3. Integrate using the standard result $\int \frac{1}{1 + x^2} , dx = \tan^{-1}(x)$.
4. Substitute $x = \sin \theta$ back into the result to get the final answer.