If `int_(0)^(pi//2) ( cos x dx)/( 6-5 sin x + sin^(2) x)= log K` then `K` is equal to

Options

A. `(2)/(3)`
B. `(4)/(3)`
C. `(1)/(3)`
D. `(5)/(3)`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{6 - 5 \sin x + \sin^2 x} \, dx \]

we will use the substitution \( \sin x = t \). This gives us \( \cos x \, dx = dt \). The limits of integration change as follows: when \( x = 0 \), \( t = \sin(0) = 0 \) and when \( x = \frac{\pi}{2} \), \( t = \sin\left(\frac{\pi}{2}\right) = 1 \).

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{1} \frac{1}{6 - 5t + t^2} \, dt \]

Next, we will simplify the denominator \( 6 - 5t + t^2 \). We can complete the square:

\[ t^2 - 5t + 6 = (t - \frac{5}{2})^2 - \frac{25}{4} + 6 = (t - \frac{5}{2})^2 - \frac{25}{4} + \frac{24}{4} = (t - \frac{5}{2})^2 - \frac{1}{4} \]

Now, we can rewrite the integral:

\[ I = \int_{0}^{1} \frac{1}{(t - \frac{5}{2})^2 - \left(\frac{1}{2}\right)^2} \, dt \]

This integral resembles the standard form:

\[ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C \]

In our case, let \( x = t - \frac{5}{2} \) and \( a = \frac{1}{2} \). Therefore, we have:

\[ I = \frac{1}{2 \cdot \frac{1}{2}} \int_{0}^{1} \frac{1}{(t - \frac{5}{2})^2 - \left(\frac{1}{2}\right)^2} \, dt = \int_{0}^{1} \frac{1}{(t - \frac{5}{2})^2 - \left(\frac{1}{2}\right)^2} \, dt \]

Now we evaluate the integral:

\[ I = \left[ \log \left| \frac{t - \frac{5}{2} - \frac{1}{2}}{t - \frac{5}{2} + \frac{1}{2}} \right| \right]_{0}^{1} \]

Calculating the limits:

1. For \( t = 1 \):

\[ \log \left| \frac{1 - \frac{5}{2} - \frac{1}{2}}{1 - \frac{5}{2} + \frac{1}{2}} \right| = \log \left| \frac{1 - 3}{1 - 2} \right| = \log \left| \frac{-2}{-1} \right| = \log 2 \]

2. For \( t = 0 \):

\[ \log \left| \frac{0 - \frac{5}{2} - \frac{1}{2}}{0 - \frac{5}{2} + \frac{1}{2}} \right| = \log \left| \frac{-3}{-2} \right| = \log \frac{3}{2} \]

Thus, we have:

\[ I = \log 2 - \log \frac{3}{2} = \log \frac{2}{\frac{3}{2}} = \log \frac{4}{3} \]

Since we are given that \( I = \log K \), we can equate:

\[ \log K = \log \frac{4}{3} \]

Therefore, we find:

\[ K = \frac{4}{3} \]

### Final Answer:
\[ K = \frac{4}{3} \]