Question Statement

Evaluate the definite integral:

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}(\sec x+\tan x{)}^{2}dx$

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Background and Explanation

This problem requires expanding a squared trigonometric expression and applying fundamental integration formulas for secant and tangent functions. You'll also need the Pythagorean identity ${\tan }^{2}x={\sec }^{2}x-1$ to simplify the integration.

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Solution

First, we expand the integrand using the algebraic identity $(a+b{)}^2=a^2+2ab+b^2$:

$(\sec x+\tan x{)}^2={\sec }^{2}x+{\tan }^{2}x+2\sec x\tan x$

Now we split the integral into three separate parts:

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}(\sec x+\tan x{)}^{2}dx={\int }_{-\pi /4}^{\pi /4}{\sec }^{2}xdx+{\int }_{-\pi /4}^{\pi /4}{\tan }^{2}xdx+2{\int }_{-\pi /4}^{\pi /4}\sec x\tan xdx$

Evaluating the first integral:

Since $\frac{d}{dx}(\tan x)={\sec }^{2}x$, we have:

${\int }_{-\pi /4}^{\pi /4}{\sec }^{2}xdx={\left[\tan x\right]}_{-\pi /4}^{\pi /4}=\tan \left(\frac{\pi }{4}\right)-\tan \left(-\frac{\pi }{4}\right)=1-(-1)=2$

Evaluating the second integral:

We use the identity ${\tan }^{2}x={\sec }^{2}x-1$ to rewrite this as:

${\int }_{-\pi /4}^{\pi /4}{\tan }^{2}xdx={\int }_{-\pi /4}^{\pi /4}({\sec }^{2}x-1)dx={\int }_{-\pi /4}^{\pi /4}{\sec }^{2}xdx-{\int }_{-\pi /4}^{\pi /4}1dx$

Computing each part:

• ${\int }_{-\pi /4}^{\pi /4}{\sec }^{2}xdx={\left[\tan x\right]}_{-\pi /4}^{\pi /4}=1-(-1)=2$
• ${\int }_{-\pi /4}^{\pi /4}1dx={\left[x\right]}_{-\pi /4}^{\pi /4}=\frac{\pi }{4}-\left(-\frac{\pi }{4}\right)=\frac{\pi }{2}$

Therefore: ${\int }_{-\pi /4}^{\pi /4}{\tan }^{2}xdx=2-\frac{\pi }{2}$

Evaluating the third integral:

Since $\frac{d}{dx}(\sec x)=\sec x\tan x$, we have:

$2{\int }_{-\pi /4}^{\pi /4}\sec x\tan xdx=2{\left[\sec x\right]}_{-\pi /4}^{\pi /4}=2\left(\sec \left(\frac{\pi }{4}\right)-\sec \left(-\frac{\pi }{4}\right)\right)$

Since $\sec (-x)=\sec (x)$ (secant is an even function), and $\sec (\frac{\pi }{4})=\sqrt{2}$:

$2\left(\sqrt{2}-\sqrt{2}\right)=0$

Combining all results:

Adding the three evaluated integrals together:

$I=2+\left(2-\frac{\pi }{2}\right)+0=4-\frac{\pi }{2}$

Thus, the final answer is:

$\boxed{4-\frac{\pi }{2}}$

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Key Formulas or Methods Used

• Algebraic expansion: $(a+b{)}^2=a^2+2ab+b^2$
• Pythagorean identity: ${\tan }^{2}x={\sec }^{2}x-1$
• Integration formulas:
  • $\int {\sec }^{2}xdx=\tan x+C$
  • $\int \sec x\tan xdx=\sec x+C$
  • $\int 1dx=x+C$
• Even function property: $\sec (-x)=\sec (x)$
• Evaluation notation: ${\left[F(x)\right]}_a^b=F(b)-F(a)$

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Summary of Steps

1. Expand the squared term $(\sec x+\tan x{)}^2$ into ${\sec }^{2}x+{\tan }^{2}x+2\sec x\tan x$
2. Split the integral into three separate definite integrals
3. Apply identity ${\tan }^{2}x={\sec }^{2}x-1$ to the second integral
4. Integrate each term:
   • ${\sec }^{2}x\to \tan x$
   • ${\tan }^{2}x\to \tan x-x$ (after identity substitution)
   • $\sec x\tan x\to \sec x$
5. Evaluate at bounds $-\frac{\pi }{4}$ and $\frac{\pi }{4}$, using $\tan (\frac{\pi }{4})=1$, $\tan (-\frac{\pi }{4})=-1$, and $\sec (\pm \frac{\pi }{4})=\sqrt{2}$
6. Combine results: $2+(2-\frac{\pi }{2})+0=4-\frac{\pi }{2}$