Key Concepts and Formulas

1. Trigonometric Identities: $\tan \theta = \frac{\sin \theta}{\cos \theta}$, $\sec \theta = \frac{1}{\cos \theta}$.
2. Substitution Method: Used to simplify integrals by changing the variable of integration. If $u = g(\theta)$, then $du = g'(\theta) \, d\theta$.
3. Power Rule for Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$).
4. Definite Integration (Fundamental Theorem of Calculus): $\int_a^b f(x) \, dx = F(b) - F(a)$, where $F(x)$ is an antiderivative of $f(x)$.
5. Algebraic Manipulation: Solving equations involving radicals and fractions.

Step-by-Step Solution

We are given the definite integral equation: $\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\quad \left( {k > 0} \right)$

Step 1: Simplify the Integrand To simplify the integrand, we express $\tan \theta$ and $\sec \theta$ in terms of $\sin \theta$ and $\cos \theta$. $\frac{\tan \theta}{\sqrt{2k\,\sec \theta}} = \frac{\frac{\sin \theta}{\cos \theta}}{\sqrt{2k \cdot \frac{1}{\cos \theta}}} = \frac{\frac{\sin \theta}{\cos \theta}}{\sqrt{2k} \cdot \frac{1}{\sqrt{\cos \theta}}}$ We can rearrange this expression: $= \frac{1}{\sqrt{2k}} \cdot \frac{\sin \theta}{\cos \theta} \cdot \sqrt{\cos \theta} = \frac{1}{\sqrt{2k}} \cdot \frac{\sin \theta}{\sqrt{\cos \theta}}$ This can be written as: $\frac{1}{\sqrt{2k}} \sin \theta (\cos \theta)^{-1/2}$

Step 2: Apply the Substitution Method to Evaluate the Indefinite Integral Let's consider the indefinite integral part: $\int \sin \theta (\cos \theta)^{-1/2} \, d\theta$. We use the substitution method. Let $u = \cos \theta$. Differentiating with respect to $\theta$, we get $du = -\sin \theta \, d\theta$. This implies $\sin \theta \, d\theta = -du$.

Now, substitute $u$ and $du$ into the integral: $\int u^{-1/2} (-du) = -\int u^{-1/2} \, du$ Using the power rule for integration $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$: $-\frac{u^{-1/2 + 1}}{-1/2 + 1} + C = -\frac{u^{1/2}}{1/2} + C = -2u^{1/2} + C$ Substituting back $u = \cos \theta$: $-2(\cos \theta)^{1/2} + C = -2\sqrt{\cos \theta} + C$

Step 3: Evaluate the Definite Integral Now we incorporate the constant factor $\frac{1}{\sqrt{2k}}$ and the limits of integration. The left side of the given equation is: $\frac{1}{\sqrt{2k}} \left[ -2\sqrt{\cos \theta} \right]_0^{\pi/3}$ Using the Fundamental Theorem of Calculus: $\frac{1}{\sqrt{2k}} \left( (-2\sqrt{\cos(\pi/3)}) - (-2\sqrt{\cos(0)}) \right)$ We know that $\cos(\pi/3) = \frac{1}{2}$ and $\cos(0) = 1$. $= \frac{1}{\sqrt{2k}} \left( (-2\sqrt{\frac{1}{2}}) - (-2\sqrt{1}) \right)$ $= \frac{1}{\sqrt{2k}} \left( (-2 \cdot \frac{1}{\sqrt{2}}) - (-2) \right)$ $= \frac{1}{\sqrt{2k}} \left( -\frac{2}{\sqrt{2}} + 2 \right) = \frac{1}{\sqrt{2k}} \left( -\sqrt{2} + 2 \right)$ $= \frac{2 - \sqrt{2}}{\sqrt{2k}}$

Step 4: Solve for k We equate this result to the right side of the given equation: $\frac{2 - \sqrt{2}}{\sqrt{2k}} = 1 - \frac{1}{\sqrt{2}}$ Let's simplify both sides. The right side can be written as: $1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$. The term $(2 - \sqrt{2})$ in the numerator of the left side can be factored as $\sqrt{2}(\sqrt{2} - 1)$. So, the left side becomes: $\frac{\sqrt{2}(\sqrt{2} - 1)}{\sqrt{2}\sqrt{k}} = \frac{\sqrt{2} - 1}{\sqrt{k}}$ Now, the equation is: $\frac{\sqrt{2} - 1}{\sqrt{k}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$ Since $k > 0$, $\sqrt{k}$ is real and positive. Also, $\sqrt{2} - 1 \neq 0$. We can divide both sides by $(\sqrt{2} - 1)$: $\frac{1}{\sqrt{k}} = \frac{1}{\sqrt{2}}$ This implies: $\sqrt{k} = \sqrt{2}$ Squaring both sides to find $k$: $k = 2$

Common Mistakes & Tips

• Sign errors during substitution: Be careful with the negative sign when $du = -\sin \theta \, d\theta$.
• Simplifying radicals: Ensure that expressions like $\sqrt{1/2}$ and $2/\sqrt{2}$ are simplified correctly.
• Algebraic manipulation of fractions and radicals: Double-check each step when solving for the unknown constant $k$.

Summary

The problem requires evaluating a definite integral involving trigonometric functions. We simplify the integrand using trigonometric identities, apply the substitution method to find the antiderivative, and then use the Fundamental Theorem of Calculus to evaluate the definite integral. Finally, we equate the result to the given value and solve for the constant $k$ using algebraic manipulation. The key steps involved transforming the integral into a solvable form and carefully handling the limits of integration.

The final answer is $\boxed{4}$.