Key Concepts and Formulas

• Trigonometric Identities:
  • $\tan x = \frac{\sin x}{\cos x}$, $\cot x = \frac{\cos x}{\sin x}$
  • $\sin^2 x + \cos^2 x = 1$
  • $\sin 2x = 2 \sin x \cos x$
• Definite Integration by Substitution: If $u = g(x)$, then $du = g'(x) dx$. The limits of integration must be changed from $x$-values to $u$-values.
• Power Rule for Integration: $\int t^n \,dt = \frac{t^{n+1}}{n+1} + C$

Step-by-Step Solution

Step 1: Simplifying the Denominator

The integrand contains the term $(\tan x + \cot x)$ in the denominator. We will simplify this using trigonometric identities. $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}$ Finding a common denominator: $\tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x}$ Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$: $\tan x + \cot x = \frac{1}{\sin x \cos x}$ Using the double angle identity for sine, $\sin 2x = 2 \sin x \cos x$, which implies $\sin x \cos x = \frac{\sin 2x}{2}$: $\tan x + \cot x = \frac{1}{\frac{\sin 2x}{2}} = \frac{2}{\sin 2x}$

• Reasoning: Simplifying the denominator into a single trigonometric function of a double angle is crucial for transforming the integrand into a form that is easier to integrate.

Step 2: Rewriting the Integral

Now, we substitute the simplified denominator back into the original integral and simplify the integrand. The original integral is: $I = \int_{{\pi \over {12}}}^{{\pi \over 4}} {\,\,{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,dx$ Substitute $\tan x + \cot x = \frac{2}{\sin 2x}$: $I = \int_{{\pi \over {12}}}^{{\pi \over 4}} {\,\,{{8\cos 2x} \over {{{\left( {{2 \over {\sin 2x}}} \right)}^3}}}} \,dx$ Simplify the denominator: $\left( \frac{2}{\sin 2x} \right)^3 = \frac{8}{\sin^3 2x}$ Substitute this back into the integral: $I = \int_{{\pi \over {12}}}^{{\pi \over 4}} {\,\,{{8\cos 2x} \over {{8 \over {\sin^3 2x}}}}} \,dx$ The '8' in the numerator and denominator cancels out, and $\sin^3 2x$ moves to the numerator: $I = \int_{{\pi \over {12}}}^{{\pi \over 4}} {\sin^3 2x \cdot \cos 2x} \,dx$

• Reasoning: This step transforms the integral into a much simpler form, $\int \sin^3 2x \cos 2x \,dx$, which is ideally suited for a substitution method.

Step 3: Applying Substitution

We will now use the substitution method to evaluate the integral $I = \int_{{\pi \over {12}}}^{{\pi \over 4}} {\sin^3 2x \cdot \cos 2x} \,dx$. Let $t = \sin 2x$. Differentiating with respect to $x$: $\frac{dt}{dx} = \frac{d}{dx}(\sin 2x) = \cos 2x \cdot 2 = 2 \cos 2x$ Rearranging, we get: $dt = 2 \cos 2x \,dx \implies \cos 2x \,dx = \frac{1}{2} dt$

• Reasoning: By choosing $t = \sin 2x$, the derivative $\cos 2x \,dx$ appears in the integrand, allowing us to replace the entire expression with a simpler form involving $t$ and $dt$.

Step 4: Changing the Limits of Integration

Since this is a definite integral, we must change the limits of integration from $x$-values to $t$-values.

• Lower limit: When $x = \frac{\pi}{12}$, $t = \sin \left( 2 \cdot \frac{\pi}{12} \right) = \sin \left( \frac{\pi}{6} \right) = \frac{1}{2}$
• Upper limit: When $x = \frac{\pi}{4}$, $t = \sin \left( 2 \cdot \frac{\pi}{4} \right) = \sin \left( \frac{\pi}{2} \right) = 1$
• Reasoning: Changing the limits ensures that we evaluate the definite integral with respect to the new variable $t$ over the correct interval, avoiding the need to substitute back for $x$.

Step 5: Evaluating the Transformed Integral

Now, substitute $t$, $dt$, and the new limits into the integral: $I = \int_{{1 \over 2}}^{1} {t^3 \cdot \left( \frac{1}{2} dt \right)}$ Factor out the constant $\frac{1}{2}$: $I = \frac{1}{2} \int_{{1 \over 2}}^{1} {t^3} \,dt$ Integrate $t^3$ using the power rule: $I = \frac{1}{2} \left[ \frac{t^4}{4} \right]_{{1 \over 2}}^{1}$ Apply the Fundamental Theorem of Calculus (Upper Limit - Lower Limit): $I = \frac{1}{2} \left( \frac{(1)^4}{4} - \frac{\left( \frac{1}{2} \right)^4}{4} \right)$

• Reasoning: This step involves performing the actual integration of the simplified polynomial in $t$ and then applying the limits to find the definite value.

Step 6: Final Calculation

Perform the arithmetic to find the final value of the integral. $I = \frac{1}{2} \left( \frac{1}{4} - \frac{\frac{1}{16}}{4} \right)$ $I = \frac{1}{2} \left( \frac{1}{4} - \frac{1}{64} \right)$ Find a common denominator to subtract the fractions: $I = \frac{1}{2} \left( \frac{16}{64} - \frac{1}{64} \right)$ $I = \frac{1}{2} \left( \frac{15}{64} \right)$ $I = \frac{15}{128}$

• Reasoning: This is the final arithmetic calculation to arrive at the numerical value of the integral.

Common Mistakes & Tips

• Trigonometric Identity Errors: Ensure accurate recall and application of trigonometric identities. A mistake here will propagate through the entire solution.
• Forgetting to Change Limits: When using substitution in definite integrals, it is crucial to change the limits of integration. If you don't, you must substitute back to the original variable before evaluating.
• Algebraic Errors with Fractions: Be meticulous with fraction arithmetic, especially when dealing with powers and subtractions.

Summary

The integral was evaluated by first simplifying the denominator using trigonometric identities to obtain $\frac{2}{\sin 2x}$. This transformed the integrand into $\sin^3 2x \cos 2x$. A substitution $t = \sin 2x$ was then employed, along with changing the limits of integration. The resulting integral in terms of $t$ was a simple power function, which was easily integrated. The final calculation yielded the value $\frac{15}{128}$.

The final answer is \boxed{\frac{15}{128}} which corresponds to option (A).