The equation `int_(-pi/4)^(pi/4){a|sinx|+(bsinx)/(1+cos^2x)+c}dx=0` where `a,b,c` are constants gives a relation between

Options

A. a, b and c
B. a and c
C. a and b
D. b and c

Answer

To solve the equation

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( a |\sin x| + \frac{b \sin x}{1 + \cos^2 x} + c \right) dx = 0, \]

where \( a, b, c \) are constants, we will break the integral into two parts based on the behavior of the sine function over the interval \([- \frac{\pi}{4}, \frac{\pi}{4}]\).

### Step 1: Break the Integral into Two Parts

The sine function is negative in the interval \([- \frac{\pi}{4}, 0]\) and positive in the interval \([0, \frac{\pi}{4}]\). Thus, we can rewrite the integral as:

\[ \int_{-\frac{\pi}{4}}^{0} \left( -a \sin x + \frac{b \sin x}{1 + \cos^2 x} + c \right) dx + \int_{0}^{\frac{\pi}{4}} \left( a \sin x + \frac{b \sin x}{1 + \cos^2 x} + c \right) dx = 0. \]

### Step 2: Evaluate Each Integral

Now we evaluate each part separately.

1. **For the first integral** from \(-\frac{\pi}{4}\) to \(0\):

\[ \int_{-\frac{\pi}{4}}^{0} \left( -a \sin x + \frac{b \sin x}{1 + \cos^2 x} + c \right) dx = -a \int_{-\frac{\pi}{4}}^{0} \sin x \, dx + b \int_{-\frac{\pi}{4}}^{0} \frac{\sin x}{1 + \cos^2 x} \, dx + c \int_{-\frac{\pi}{4}}^{0} dx. \]

2. **For the second integral** from \(0\) to \(\frac{\pi}{4}\):

\[ \int_{0}^{\frac{\pi}{4}} \left( a \sin x + \frac{b \sin x}{1 + \cos^2 x} + c \right) dx = a \int_{0}^{\frac{\pi}{4}} \sin x \, dx + b \int_{0}^{\frac{\pi}{4}} \frac{\sin x}{1 + \cos^2 x} \, dx + c \int_{0}^{\frac{\pi}{4}} dx. \]

### Step 3: Combine the Results

Combining the two integrals, we have:

\[ -a \left[ -\cos x \right]_{-\frac{\pi}{4}}^{0} + b \int_{-\frac{\pi}{4}}^{0} \frac{\sin x}{1 + \cos^2 x} \, dx + c \left[ x \right]_{-\frac{\pi}{4}}^{0} + a \left[ -\cos x \right]_{0}^{\frac{\pi}{4}} + b \int_{0}^{\frac{\pi}{4}} \frac{\sin x}{1 + \cos^2 x} \, dx + c \left[ x \right]_{0}^{\frac{\pi}{4}} = 0. \]

### Step 4: Evaluate the Limits

1. Evaluating the limits for \(-a \left[ -\cos x \right]_{-\frac{\pi}{4}}^{0}\):

\[ -a \left( -\cos(0) + \cos\left(-\frac{\pi}{4}\right) \right) = -a \left( -1 + \frac{1}{\sqrt{2}} \right) = a \left( 1 - \frac{1}{\sqrt{2}} \right). \]

2. Evaluating the limits for \(a \left[ -\cos x \right]_{0}^{\frac{\pi}{4}}\):

\[ a \left( -\cos\left(\frac{\pi}{4}\right) + \cos(0) \right) = a \left( -\frac{1}{\sqrt{2}} + 1 \right) = a \left( 1 - \frac{1}{\sqrt{2}} \right). \]

3. Evaluating the constant terms:

\[ c \left[ x \right]_{-\frac{\pi}{4}}^{0} = c \left( 0 + \frac{\pi}{4} \right) = \frac{c \pi}{4}, \]
\[ c \left[ x \right]_{0}^{\frac{\pi}{4}} = c \left( \frac{\pi}{4} - 0 \right) = \frac{c \pi}{4}. \]

### Step 5: Combine Everything to Form the Equation

Combining all parts, we have:

\[ a \left( 1 - \frac{1}{\sqrt{2}} \right) + a \left( 1 - \frac{1}{\sqrt{2}} \right) + b \left( \text{integrals} \right) + \frac{c \pi}{2} = 0. \]

### Step 6: Final Relation

This leads to the relation:

\[ 2a \left( 1 - \frac{1}{\sqrt{2}} \right) + b \left( \text{integrals} \right) + \frac{c \pi}{2} = 0. \]

This is the required relation between \(a\), \(b\), and \(c\).