If ∫ (1/x) √((1-x)/(1+x)) dx = g(x) + c, g(1) = 0, then g(1/2) is equal to : (A) log_e((√3-1)/(√3+1)

Q: Why did we choose $x = \cos 2\theta$ instead of $x = \cos \theta$?

Great question! Using $x = \cos 2\theta$ allows us to use the identities $1 - \cos 2\theta = 2 \sin^2 \theta$ and $1 + \cos 2\theta = 2 \cos^2 \theta$. This simplifies the radical to $\tan \theta$ perfectly. If you used $x = \cos \theta$, you would be stuck with half-angles like $\theta/2$, which makes the subsequent integration much messier.

Q: Why does $-\ln |\sec 2\theta + \tan 2\theta|$ become $\ln |\sec 2\theta - \tan 2\theta|$?

This is a beautiful property of trigonometry! Note that $(\sec 2\theta + \tan 2\theta)(\sec 2\theta - \tan 2\theta) = \sec^2 2\theta - \tan^2 2\theta = 1$. Therefore, $\sec 2\theta + \tan 2\theta = \frac{1}{\sec 2\theta - \tan 2\theta}$. Taking the negative log of the reciprocal is the same as the positive log of the original value.

Q: How do we handle the constant $c$?

The problem gives us the condition $g(1) = 0$. Once you have your expression for $g(x)$, simply plug in $x=1$ and solve for $c$. In many indefinite integral problems, the constant is determined by such boundary conditions.

Question

If $\int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} dx = g(x) + c, g(1) = 0$, then $g\left(\frac{1}{2}\right)$ is equal to :

Options:

(A) $\log_e \left( \frac{\sqrt{3}-1}{\sqrt{3}+1} \right) + \frac{\pi}{3}$

(B) $\log_e \left( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right) + \frac{\pi}{3}$

(C) $\log_e \left( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right) - \frac{\pi}{3}$

(D) $\frac{1}{2} \log_e \left( \frac{\sqrt{3}-1}{\sqrt{3}+1} \right) - \frac{\pi}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the integral:

$$I = \int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} \, dx = g(x) + c$$

given the boundary condition $g(1) = 0$. The radical term $\sqrt{\frac{1-x}{1+x}}$ is a classic indicator for a trigonometric substitution.

### Phase 1: The Strategic Substitution

We choose the substitution $x = \cos 2	heta$. This allows us to invoke the half-angle identities:

$$1 - \cos 2	heta = 2 \sin^2 	heta \quad 	ext{and} \quad 1 + \cos 2	heta = 2 \cos^2 	heta$$

Substituting these into the radical, we obtain:

$$\sqrt{\frac{1-\cos 2	heta}{1+\cos 2	heta}} = \sqrt{\frac{2 \sin^2 	heta}{2 \cos^2 	heta}} = 	an 	heta$$

Next, we transform the differential $dx$. Differentiating $x = \cos 2	heta$ with respect to $	heta$ yields:

$$dx = -2 \sin 2	heta \, d	heta$$

### Phase 2: The Algebraic Dance

Assembling the components, the integral becomes:

$$I = \int \frac{1}{\cos 2	heta} (	an 	heta) (-2 \sin 2	heta) \, d	heta$$

Using the identities $\sin 2	heta = 2 \sin 	heta \cos 	heta$ and $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$, the expression simplifies:

$$I = \int \frac{1}{\cos 2	heta} \left( \frac{\sin 	heta}{\cos 	heta} ight) (-4 \sin 	heta \cos 	heta) \, d	heta = \int \frac{-4 \sin^2 	heta}{\cos 2	heta} \, d	heta$$

Applying $2 \sin^2 	heta = 1 - \cos 2	heta$, we transform the integral into:

$$I = -2 \int \left( \frac{1 - \cos 2	heta}{\cos 2	heta} ight) \, d	heta = -2 \int (\sec 2	heta - 1) \, d	heta$$

### Phase 3: The Final Integration

Integrating the terms, we get:

$$I = -2 \left( \frac{1}{2} \ln |\sec 2	heta + 	an 2	heta| - 	heta ight) + c = -\ln |\sec 2	heta + 	an 2	heta| + 2	heta + c$$

Using the property $-\ln|u| = \ln|1/u|$, we rewrite this as:

$$g(x) = \ln |\sec 2	heta - 	an 2	heta| + 2	heta + c$$

Since $\cos 2	heta = x$, we have $\sec 2	heta = \frac{1}{x}$ and $	an 2	heta = \frac{\sqrt{1-x^2}}{x}$. Substituting these back:

$$g(x) = \ln \left| \frac{1 - \sqrt{1-x^2}}{x} ight| + \cos^{-1} x + c$$

### Final Calculation

Applying the condition $g(1) = 0$:

$$g(1) = \ln |1 - 0| + \cos^{-1}(1) + c = 0 + 0 + c = 0 \Rightarrow c = 0$$

Evaluating at $x = \frac{1}{2}$:

$$g\left(\frac{1}{2}ight) = \ln \left| \frac{1 - \sqrt{3/4}}{1/2} ight| + \cos^{-1}\left(\frac{1}{2}ight) = \ln |2 - \sqrt{3}| + \frac{\pi}{3}$$

The final result is **$\ln(2 - \sqrt{3}) + \frac{\pi}{3}$**.

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{1}{x} \frac{1 - x}{1 + x} d x = g (x) + c, g (1) = 0$ , then $g (\frac{1}{2})$ is equal to :

Select Answer:

Visualized Solution

Problem Analysis

Choosing the Substitution

Finding the Differential

Substituting into the Integral

Simplifying the Radicals

Using Double Angle Identities

Reducing the Integrand

Splitting the Integral

Performing the Integration

Transforming the Logarithm

Returning to Variable $x$

Evaluating at $x = \frac{1}{2}$

Final Simplification

The Sigma Insight: Integration by Substitution

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

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