3.6 Q-27

Question Statement

Evaluate the following definite integral:

$$\int_{0}^{\pi / 4} \frac{1}{1 + \sin x} , dx$$

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Background and Explanation

To solve this problem, we need to use a trigonometric identity and break down the integrand. The key concepts include:

• Secant and Tangent functions and their integration formulas.
• The identity:

$$1 - \sin^2 x = \cos^2 x$$

This will help us manipulate the given expression to a form that we can easily integrate.

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Solution

Step 1: Multiply by a conjugate

To simplify the integrand, we multiply both the numerator and denominator by the conjugate of the denominator:

$$\frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x}$$

This gives:

$$\int_{0}^{\pi / 4} \frac{1 - \sin x}{(1 - \sin^2 x)} , dx$$

Using the identity $1 - \sin^2 x = \cos^2 x$, we rewrite the denominator:

$$\int_{0}^{\pi / 4} \frac{1 - \sin x}{\cos^2 x} , dx$$

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Step 2: Split the integrand

Now, we split the integrand into two separate terms:

$$\int_{0}^{\pi / 4} \frac{1}{\cos^2 x} , dx - \int_{0}^{\pi / 4} \frac{\sin x}{\cos^2 x} , dx$$

This simplifies to:

$$\int_{0}^{\pi / 4} \sec^2 x , dx - \int_{0}^{\pi / 4} \sec x \tan x , dx$$

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Step 3: Integrate each term

• First integral: The integral of $\sec^2 x$ is:

$$\int \sec^2 x , dx = \tan x$$

Evaluating from 0 to $\frac{\pi}{4}$:

$$\tan\left(\frac{\pi}{4}\right) - \tan(0)$$

Since $\tan\left(\frac{\pi}{4}\right) = 1$ and $\tan(0) = 0$, this gives:

$$1 - 0 = 1$$

• Second integral: The integral of $\sec x \tan x$ is:

$$\int \sec x \tan x , dx = \sec x$$

Evaluating from 0 to $\frac{\pi}{4}$:

$$\sec\left(\frac{\pi}{4}\right) - \sec(0)$$

Since $\sec\left(\frac{\pi}{4}\right) = \sqrt{2}$ and $\sec(0) = 1$, this gives:

$$\sqrt{2} - 1$$

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Step 4: Combine the results

Now, we combine the results from both integrals:

$$1 - (\sqrt{2} - 1) = 1 - \sqrt{2} + 1 = 2 - \sqrt{2}$$

Thus, the value of the integral is:

$$\boxed{2 - \sqrt{2}}$$

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Key Formulas or Methods Used

• Integration of secant and tangent:
  • $\int \sec^2 x , dx = \tan x$
  • $\int \sec x \tan x , dx = \sec x$
• Trigonometric identity:
  • $1 - \sin^2 x = \cos^2 x$

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Summary of Steps

1. Multiply the integrand by the conjugate of $1 + \sin x$: $\frac{1 - \sin x}{\cos^2 x}$

2. Split the integrand into two separate integrals: $\int_{0}^{\pi / 4} \sec^2 x , dx - \int_{0}^{\pi / 4} \sec x \tan x , dx$

3. Integrate each term:

   • $\int \sec^2 x , dx = \tan x$
   • $\int \sec x \tan x , dx = \sec x$

4. Evaluate the integrals at the limits $0$ and $\frac{\pi}{4}$.

5. Combine the results to get the final answer: $2 - \sqrt{2}$

The final answer is:

$$\boxed{2 - \sqrt{2}}$$