prove that integration of root (x^2-a^2) dx=x/2root(x^2-a^2)-a^2/2 lnMo x-root(x^2-a^2) +c where x=asecθ

To prove that the integral of \(\sqrt{x^2 - a^2} \, dx\) is equal to the given result using the substitution \( x = a \sec \theta \), follow the steps below:

### Problem:
\[
I = \int \sqrt{x^2 - a^2} \, dx
\]

### Substitution:
Let \( x = a \sec \theta \). Then:

\[
dx = a \sec \theta \tan \theta \, d\theta
\]

Also, from the identity \( \sec^2 \theta - 1 = \tan^2 \theta \), we get:

\[
x^2 - a^2 = a^2 \sec^2 \theta - a^2 = a^2 (\sec^2 \theta - 1) = a^2 \tan^2 \theta
\]

Thus:

\[
\sqrt{x^2 - a^2} = \sqrt{a^2 \tan^2 \theta} = a \tan \theta
\]

Now, substitute into the integral:

\[
I = \int \sqrt{x^2 - a^2} \, dx = \int a \tan \theta \cdot a \sec \theta \tan \theta \, d\theta
\]

\[
I = a^2 \int \tan^2 \theta \sec \theta \, d\theta
\]

### Simplify:
We use the identity \( \tan^2 \theta = \sec^2 \theta - 1 \) to rewrite the integral:

\[
I = a^2 \int (\sec^2 \theta - 1) \sec \theta \, d\theta
\]

Expanding:

\[
I = a^2 \int (\sec^3 \theta - \sec \theta) \, d\theta
\]

Now, we solve this integral in parts.

### Solving the Integral:

#### 1. Integral of \( \sec^3 \theta \):
We use the known result for \( \int \sec^3 \theta \, d\theta \):

\[
\int \sec^3 \theta \, d\theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| + C
\]

#### 2. Integral of \( \sec \theta \):
The integral of \( \sec \theta \) is:

\[
\int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta| + C
\]

### Combining the Results:
Now, substitute these results into the original integral:

\[
I = a^2 \left( \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| - \ln |\sec \theta + \tan \theta| \right) + C
\]

Simplify:

\[
I = a^2 \left( \frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln |\sec \theta + \tan \theta| \right) + C
\]

### Substitute Back \( x = a \sec \theta \):
Now, express everything back in terms of \( x \). Recall that \( \sec \theta = \frac{x}{a} \) and \( \tan \theta = \sqrt{\frac{x^2 - a^2}{a^2}} = \frac{\sqrt{x^2 - a^2}}{a} \).

So:

\[
\sec \theta \tan \theta = \frac{x}{a} \cdot \frac{\sqrt{x^2 - a^2}}{a} = \frac{x \sqrt{x^2 - a^2}}{a^2}
\]

And:

\[
\sec \theta + \tan \theta = \frac{x}{a} + \frac{\sqrt{x^2 - a^2}}{a}
\]

Thus:

\[
\ln |\sec \theta + \tan \theta| = \ln \left| \frac{x + \sqrt{x^2 - a^2}}{a} \right| = \ln |x + \sqrt{x^2 - a^2}| - \ln a
\]

Now, substituting back:

\[
I = \frac{x \sqrt{x^2 - a^2}}{2} - \frac{a^2}{2} \ln |x + \sqrt{x^2 - a^2}| + C
\]

This completes the proof.

### Final Answer:
\[
\int \sqrt{x^2 - a^2} \, dx = \frac{x \sqrt{x^2 - a^2}}{2} - \frac{a^2}{2} \ln |x + \sqrt{x^2 - a^2}| + C
\]

Would you like more details or have any questions?

#### 5 Related Questions:
1. How do you handle integrals involving other trigonometric substitutions?
2. Why is the identity \( \sec^2 \theta - 1 = \tan^2 \theta \) useful in trigonometric substitution?
3. What are the typical applications of trigonometric substitution in calculus?
4. How do we handle different cases of integrals with square roots like \( \sqrt{a^2 - x^2} \) or \( \sqrt{a^2 + x^2} \)?
5. Can this method be extended to solve integrals with higher powers of \( x \) in the square root?

#### Tip:
Trigonometric substitutions work best when you have quadratic terms under a square root or within denominators, often simplifying complex expressions into basic trigonometric integrals.

Prove that the integration of √(x² - a²) dx = (x/2)√(x² - a²) - (a²/2) ln|x - √(x² - a²)| + C where x = a sec θ

Learn how to solve the integral of √(x² - a²) using trigonometric substitution, specifically x = a sec θ. This problem explores key calculus concepts, including integration by substitution, and involves advanced trigonometric identities. Suitable for advanced high school and university students.

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