If the integral `I=int(tanx)/(5+7tan^(2)x)dx=kln |f(x)|+C` (where C is the integration constant) and `f(0)=5/7`, then the value of `f((pi)/(4))` is equal to

Answer

To solve the integral

\[ I = \int \frac{\tan x}{5 + 7 \tan^2 x} \, dx = k \ln |f(x)| + C \]

where \( f(0) = \frac{5}{7} \), we need to find the value of \( f\left(\frac{\pi}{4}\right) \).

### Step 1: Substitution
Let \( t = 5 + 7 \tan^2 x \). Then, we differentiate \( t \):

\[ \frac{dt}{dx} = 14 \tan x \sec^2 x \implies dt = 14 \tan x \sec^2 x \, dx \]

From this, we can express \( \tan x \, dx \):

\[ \tan x \, dx = \frac{dt}{14 \sec^2 x} \]

### Step 2: Express \(\sec^2 x\) in terms of \(t\)
Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can express \( \tan^2 x \) in terms of \( t \):

\[ \tan^2 x = \frac{t - 5}{7} \implies \sec^2 x = 1 + \frac{t - 5}{7} = \frac{t + 2}{7} \]

### Step 3: Substitute into the integral
Now substituting back into the integral, we have:

\[ I = \int \frac{\tan x}{t} \cdot \frac{dt}{14 \sec^2 x} \]

Substituting for \(\sec^2 x\):

\[ I = \int \frac{\tan x}{t} \cdot \frac{dt}{14 \cdot \frac{t + 2}{7}} = \int \frac{7 \tan x}{14t(t + 2)} \, dt = \frac{1}{2} \int \frac{\tan x}{t(t + 2)} \, dt \]

### Step 4: Partial Fraction Decomposition
We can perform partial fraction decomposition on \(\frac{1}{t(t + 2)}\):

\[ \frac{1}{t(t + 2)} = \frac{A}{t} + \frac{B}{t + 2} \]

Multiplying through by \(t(t + 2)\) gives:

\[ 1 = A(t + 2) + Bt \]

Setting \(t = 0\):

\[ 1 = 2A \implies A = \frac{1}{2} \]

Setting \(t = -2\):

\[ 1 = -2B \implies B = -\frac{1}{2} \]

Thus, we have:

\[ \frac{1}{t(t + 2)} = \frac{1/2}{t} - \frac{1/2}{t + 2} \]

### Step 5: Integrate
Now we can integrate:

\[ I = \frac{1}{2} \left( \frac{1}{2} \ln |t| - \frac{1}{2} \ln |t + 2| \right) + C \]

This simplifies to:

\[ I = \frac{1}{4} \ln \left| \frac{t}{t + 2} \right| + C \]

### Step 6: Substitute back for \(t\)
Substituting back for \(t\):

\[ I = \frac{1}{4} \ln \left| \frac{5 + 7 \tan^2 x}{7 + 7 \tan^2 x} \right| + C \]

### Step 7: Find \(f(x)\)
Comparing with \(I = k \ln |f(x)| + C\), we find:

\[ f(x) = \frac{5 + 7 \tan^2 x}{7 + 7 \tan^2 x} \]

### Step 8: Evaluate \(f(0)\)
We know \(f(0) = \frac{5 + 7 \cdot 0}{7 + 7 \cdot 0} = \frac{5}{7}\), which is consistent with the given condition.

### Step 9: Evaluate \(f\left(\frac{\pi}{4}\right)\)
Now we evaluate \(f\left(\frac{\pi}{4}\right)\):

\[ \tan\left(\frac{\pi}{4}\right) = 1 \implies f\left(\frac{\pi}{4}\right) = \frac{5 + 7 \cdot 1^2}{7 + 7 \cdot 1^2} = \frac{5 + 7}{7 + 7} = \frac{12}{14} = \frac{6}{7} \]

Thus, the value of \(f\left(\frac{\pi}{4}\right)\) is

\[ \boxed{\frac{6}{7}} \]