Let l=`int_(0)^(pi//2)(cosx)/(a cos x+ b sinx)dx` and `J=int_(0)^(pi//2)(sin x)/(a cosx+b sin x )dx`, where a ` gt 0 and b gt 0 ` Compute the values of l and j

Answer

To solve the given integrals \( I \) and \( J \), we start with the definitions:

\[ I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{a \cos x + b \sin x} \, dx \]

\[ J = \int_0^{\frac{\pi}{2}} \frac{\sin x}{a \cos x + b \sin x} \, dx \]

### Step 1: Multiply \( I \) by \( a \) and \( J \) by \( b \) and add them.

We compute \( aI + bJ \):

\[ aI = a \int_0^{\frac{\pi}{2}} \frac{\cos x}{a \cos x + b \sin x} \, dx = \int_0^{\frac{\pi}{2}} \frac{a \cos x}{a \cos x + b \sin x} \, dx \]

\[ bJ = b \int_0^{\frac{\pi}{2}} \frac{\sin x}{a \cos x + b \sin x} \, dx = \int_0^{\frac{\pi}{2}} \frac{b \sin x}{a \cos x + b \sin x} \, dx \]

Adding these two results gives:

\[ aI + bJ = \int_0^{\frac{\pi}{2}} \frac{a \cos x + b \sin x}{a \cos x + b \sin x} \, dx \]

### Step 2: Simplify the integral.

Since the numerator and denominator are the same, we have:

\[ aI + bJ = \int_0^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \]

This gives us our first equation:

\[ aI + bJ = \frac{\pi}{2} \quad \text{(Equation 1)} \]

### Step 3: Compute \( bI - aJ \).

Now we compute \( bI - aJ \):

\[ bI = b \int_0^{\frac{\pi}{2}} \frac{\cos x}{a \cos x + b \sin x} \, dx = \int_0^{\frac{\pi}{2}} \frac{b \cos x}{a \cos x + b \sin x} \, dx \]

\[ -aJ = -a \int_0^{\frac{\pi}{2}} \frac{\sin x}{a \cos x + b \sin x} \, dx = -\int_0^{\frac{\pi}{2}} \frac{a \sin x}{a \cos x + b \sin x} \, dx \]

Combining these gives:

\[ bI - aJ = \int_0^{\frac{\pi}{2}} \frac{b \cos x - a \sin x}{a \cos x + b \sin x} \, dx \]

### Step 4: Use substitution to simplify.

Let \( f(x) = a \cos x + b \sin x \). Then, the derivative \( f'(x) = -a \sin x + b \cos x \). We can rewrite the integral as:

\[ bI - aJ = \int_0^{\frac{\pi}{2}} \frac{f'(x)}{f(x)} \, dx \]

Using the property of logarithmic integration, we have:

\[ \int \frac{f'(x)}{f(x)} \, dx = \log |f(x)| \]

Thus,

\[ bI - aJ = \left[ \log(a \cos x + b \sin x) \right]_0^{\frac{\pi}{2}} = \log(b) - \log(a) = \log\left(\frac{b}{a}\right) \]

This gives us our second equation:

\[ bI - aJ = \log\left(\frac{b}{a}\right) \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations.

Now we have two equations:

1. \( aI + bJ = \frac{\pi}{2} \)
2. \( bI - aJ = \log\left(\frac{b}{a}\right) \)

We can solve these equations simultaneously. Let's express \( I \) and \( J \):

From Equation 1, we can express \( J \):

\[ J = \frac{\frac{\pi}{2} - aI}{b} \]

Substituting into Equation 2:

\[ bI - a\left(\frac{\frac{\pi}{2} - aI}{b}\right) = \log\left(\frac{b}{a}\right) \]

Multiplying through by \( b \):

\[ b^2 I - a\left(\frac{\pi}{2} - aI\right) = b \log\left(\frac{b}{a}\right) \]

Rearranging gives:

\[ (b^2 + a^2)I = \frac{a\pi}{2} + b \log\left(\frac{b}{a}\right) \]

Thus,

\[ I = \frac{1}{a^2 + b^2} \left( a \frac{\pi}{2} + b \log\left(\frac{b}{a}\right) \right) \]

Now substituting \( I \) back into Equation 1 to find \( J \):

\[ J = \frac{1}{b} \left( \frac{\pi}{2} - aI \right) \]

After substituting \( I \):

\[ J = \frac{1}{b} \left( \frac{\pi}{2} - a \cdot \frac{1}{a^2 + b^2} \left( a \frac{\pi}{2} + b \log\left(\frac{b}{a}\right) \right) \right) \]

### Final Values

After simplifying, we find:

\[ I = \frac{1}{a^2 + b^2} \left( a \frac{\pi}{2} + b \log\left(\frac{b}{a}\right) \right) \]

\[ J = \frac{1}{a^2 + b^2} \left( b \frac{\pi}{2} - a \log\left(\frac{b}{a}\right) \right) \]