Question Statement

Evaluate the definite integral:

${\int }_0^{\frac{\pi }{4}}\frac{1}{1-\sin x}dx$

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Background and Explanation

This problem involves integrating a rational trigonometric function where the denominator contains a single sine term. The standard approach is to rationalize the denominator by multiplying by the conjugate $(1+\sin x)$, which converts the expression into terms of $\sec x$ and $\tan x$ that have straightforward antiderivatives.

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Solution

To evaluate this integral, we use the technique of rationalizing the denominator to eliminate the problematic $1-\sin x$ term in the denominator.

Step 1: Rationalize the integrand

Multiply the numerator and denominator by the conjugate $(1+\sin x)$:

$\begin{array}{rl}I& ={\int }_0^{\pi /4}\frac{1}{1-\sin x}\times \frac{1+\sin x}{1+\sin x}dx\\ & ={\int }_0^{\pi /4}\frac{1+\sin x}{1-{\sin }^{2}x}dx\end{array}$

Step 2: Apply the Pythagorean identity

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

$I={\int }_0^{\pi /4}\frac{1+\sin x}{{\cos }^{2}x}dx$

Step 3: Split the fraction and convert to secant/tangent form

Separate the fraction into two terms and express them using $\sec x$ and $\tan x$:

$\begin{array}{rl}I& ={\int }_0^{\pi /4}\left(\frac{1}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{2}x}\right)dx\\ & ={\int }_0^{\pi /4}\left({\sec }^{2}x+\sec x\tan x\right)dx\end{array}$

Note that $\frac{1}{{\cos }^{2}x}={\sec }^{2}x$ and $\frac{\sin x}{{\cos }^{2}x}=\frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}=\sec x\tan x$.

Step 4: Integrate term by term

Split the integral and apply standard integration formulas:

$I={\int }_0^{\pi /4}{\sec }^{2}xdx+{\int }_0^{\pi /4}\sec x\tan xdx$

Using the standard results $\int {\sec }^{2}xdx=\tan x$ and $\int \sec x\tan xdx=\sec x$:

$I={\left[\tan x\right]}_0^{\pi /4}+{\left[\sec x\right]}_0^{\pi /4}$

Step 5: Evaluate the definite integral

Substitute the upper and lower limits:

$\begin{array}{rl}I& =\left(\tan \frac{\pi }{4}-\tan 0\right)+\left(\sec \frac{\pi }{4}-\sec 0\right)\\ & =(1-0)+(\sqrt{2}-1)\\ & =1-0+\sqrt{2}-1\\ & =\sqrt{2}\end{array}$

Therefore, the value of the integral is $\sqrt{2}$.

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Key Formulas or Methods Used

• Rationalization technique: $\frac{1}{1-\sin x}\cdot \frac{1+\sin x}{1+\sin x}=\frac{1+\sin x}{1-{\sin }^{2}x}$
• Pythagorean identity: $1-{\sin }^{2}x={\cos }^{2}x$
• Trigonometric conversions: $\frac{1}{{\cos }^{2}x}={\sec }^{2}x$ and $\frac{\sin x}{{\cos }^{2}x}=\sec x\tan x$
• Standard integrals:
  • $\int {\sec }^{2}xdx=\tan x+C$
  • $\int \sec x\tan xdx=\sec x+C$
• Evaluation values: $\tan (\pi /4)=1$, $\tan (0)=0$, $\sec (\pi /4)=\sqrt{2}$, $\sec (0)=1$

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Summary of Steps

1. Rationalize: Multiply numerator and denominator by $(1+\sin x)$ to eliminate the radical-like denominator
2. Simplify: Apply the identity $1-{\sin }^{2}x={\cos }^{2}x$ to the denominator
3. Decompose: Split $\frac{1+\sin x}{{\cos }^{2}x}$ into ${\sec }^{2}x+\sec x\tan x$
4. Integrate: Apply standard formulas to get $\tan x+\sec x$ evaluated from $0$ to $\pi /4$
5. Calculate: Substitute bounds: $(1+\sqrt{2})-(0+1)=\sqrt{2}$