Key Concepts and Formulas

• Trigonometric Identity: $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$
• Trigonometric Identity: $1 - \cos(x) = 2\sin^2(\frac{x}{2})$ and $1 + \cos(x) = 2\cos^2(\frac{x}{2})$
• Integral of tangent: $\int \tan(x) \, dx = \ln|\sec(x)| + C$
• Chain Rule for Integration (Substitution): $\int f(g(x))g'(x) \, dx = \int f(u) \, du$, where $u = g(x)$.

Step-by-Step Solution

Step 1: Rewrite the integrand using the double angle formula.

We are given the integral: $I = \int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx}$ Using the double angle formula $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, we rewrite the integrand: $I = \int {x\sqrt {{{2\sin ({x^2} - 1) - 2\sin ({x^2} - 1)\cos ({x^2} - 1)} \over {2\sin ({x^2} - 1) + 2\sin ({x^2} - 1)\cos ({x^2} - 1)}}} dx}$

Step 2: Simplify the expression inside the square root.

We factor out $2\sin(x^2 - 1)$ from the numerator and the denominator: $I = \int {x\sqrt {{{2\sin ({x^2} - 1)(1 - \cos ({x^2} - 1))} \over {2\sin ({x^2} - 1)(1 + \cos ({x^2} - 1))}}} dx}$ Canceling the common factor, we get: $I = \int {x\sqrt {{{1 - \cos ({x^2} - 1)} \over {1 + \cos ({x^2} - 1)}}} dx}$

Step 3: Use half-angle identities to further simplify.

We use the identities $1 - \cos(x) = 2\sin^2(\frac{x}{2})$ and $1 + \cos(x) = 2\cos^2(\frac{x}{2})$: $I = \int {x\sqrt {{{2\sin^2 ({{{x^2} - 1} \over 2})} \over {2\cos^2 ({{{x^2} - 1} \over 2})}}} dx}$ $I = \int {x\sqrt {{\tan^2 ({{{x^2} - 1} \over 2})}} dx}$ $I = \int {x\left| {\tan \left( {{{{x^2} - 1} \over 2}} \right)} \right|dx}$ Since we are dealing with indefinite integrals, we can assume $\tan \left( {{{{x^2} - 1} \over 2}} \right)$ is positive. Thus, $I = \int {x\tan \left( {{{{x^2} - 1} \over 2}} \right)dx}$

Step 4: Use substitution to solve the integral.

Let $t = \frac{x^2 - 1}{2}$. Then, $\frac{dt}{dx} = \frac{2x}{2} = x$, which implies $dt = x\, dx$. Substituting, we have: $I = \int {\tan (t)dt}$

Step 5: Evaluate the integral.

We know that $\int \tan(t) \, dt = \ln|\sec(t)| + C$. Therefore, $I = \ln|\sec(t)| + C$

Step 6: Substitute back to get the final answer.

Substituting $t = \frac{x^2 - 1}{2}$ back into the expression, we have: $I = \ln\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C$

However, the correct answer is given as ${\log _e}\left| {{1 \over 2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c$. Let's manipulate our answer to match the given correct answer.

Let's analyze the given correct answer: ${\log _e}\left| {{1 \over 2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c = \ln \left| \frac{1}{2} \sec^2(x^2 - 1) \right| + c = \ln \left| \frac{1}{2} \frac{1}{\cos^2(x^2 - 1)} \right| + c$ $= \ln \left| \frac{1}{2} \frac{1}{\cos^2(x^2 - 1)} \right| + c = \ln \left| \frac{1}{2} \right| + \ln \left| \sec^2(x^2 - 1) \right| + c = \ln \left| \frac{1}{2} \right| + 2\ln \left| \sec(x^2 - 1) \right| + c$

There seems to be an error in the provided "Correct Answer". Let's review our steps.

We have: $I = \int {x\tan \left( {{{{x^2} - 1} \over 2}} \right)dx}$ Let $t = \frac{x^2 - 1}{2}$. Then, $dt = x\, dx$. Substituting, we have: $I = \int {\tan (t)dt} = \ln|\sec(t)| + C = \ln \left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C$

Now, let's try to find an error in the problem statement or options.

Common Mistakes & Tips

• Remember to use the correct trigonometric identities.
• Don't forget the constant of integration.
• Double-check the substitution to ensure it is done correctly.
• Be careful about signs and simplifying the expression.

Summary

We simplified the integrand using trigonometric identities and then used substitution to evaluate the integral. The final result is $\ln\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C$. Given the ground truth, there must be an error in the options. However, based on our calculation, the most probable answer is (D).

Final Answer

The final answer is \boxed{{\log _e}\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + c}, which corresponds to option (D).