The value of the integral `int_(-pi//3)^(pi//3) (x sinx)/(cos^(2)x)dx`, is

Options

A. `(pi//3-logtan3pi//2)`
B. `2(2pi//3-logtan5pi//12)`
C. `3(pi//2-logsinpi//12)`
D. none of these

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{x \sin x}{\cos^2 x} \, dx, \]

we can start by rewriting the integrand. Notice that

\[ \frac{\sin x}{\cos^2 x} = \tan x \sec x. \]

Thus, we can express the integral as:

\[ I = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} x \tan x \sec x \, dx. \]

Next, we will use integration by parts. We set:

- \( u = x \) (which implies \( du = dx \)),
- \( dv = \tan x \sec x \, dx \).

To find \( v \), we need to integrate \( \tan x \sec x \):

\[ v = \int \tan x \sec x \, dx = \sec x. \]

Now, applying integration by parts, we have:

\[ I = \left[ x \sec x \right]_{-\frac{\pi}{3}}^{\frac{\pi}{3}} - \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \sec x \, dx. \]

Calculating the first term:

1. **Upper limit**: When \( x = \frac{\pi}{3} \):
\[ \sec\left(\frac{\pi}{3}\right) = 2 \quad \text{(since } \sec x = \frac{1}{\cos x}\text{)} \]
Thus,
\[ \frac{\pi}{3} \cdot 2 = \frac{2\pi}{3}. \]

2. **Lower limit**: When \( x = -\frac{\pi}{3} \):
\[ \sec\left(-\frac{\pi}{3}\right) = 2. \]
Thus,
\[ -\left(-\frac{\pi}{3} \cdot 2\right) = \frac{2\pi}{3}. \]

Putting it all together, we have:

\[ \left[ x \sec x \right]_{-\frac{\pi}{3}}^{\frac{\pi}{3}} = \frac{2\pi}{3} - \left(-\frac{2\pi}{3}\right) = \frac{2\pi}{3} + \frac{2\pi}{3} = \frac{4\pi}{3}. \]

Next, we need to evaluate the integral \( \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \sec x \, dx \).

Using the property of the secant function, we know:

\[ \int \sec x \, dx = \log |\sec x + \tan x| + C. \]

Thus, we evaluate:

\[ \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \sec x \, dx = \left[ \log |\sec x + \tan x| \right]_{-\frac{\pi}{3}}^{\frac{\pi}{3}}. \]

Calculating the limits:

1. **Upper limit**: When \( x = \frac{\pi}{3} \):
\[ \sec\left(\frac{\pi}{3}\right) = 2, \quad \tan\left(\frac{\pi}{3}\right) = \sqrt{3}. \]
Thus,
\[ \log |2 + \sqrt{3}|. \]

2. **Lower limit**: When \( x = -\frac{\pi}{3} \):
\[ \sec\left(-\frac{\pi}{3}\right) = 2, \quad \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}. \]
Thus,
\[ \log |2 - \sqrt{3}|. \]

Putting it together, we have:

\[ \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \sec x \, dx = \log |2 + \sqrt{3}| - \log |2 - \sqrt{3}| = \log \left(\frac{2 + \sqrt{3}}{2 - \sqrt{3}}\right). \]

Now substituting back into our expression for \( I \):

\[ I = \frac{4\pi}{3} - \log \left(\frac{2 + \sqrt{3}}{2 - \sqrt{3}}\right). \]

To simplify this further, we can rationalize the logarithm:

\[ \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{(2 + \sqrt{3})^2}{1} = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}. \]

Thus, we have:

\[ I = \frac{4\pi}{3} - \log(7 + 4\sqrt{3}). \]

Final answer:

\[ I = \frac{4\pi}{3} - \log(7 + 4\sqrt{3}). \]