The value of `int_(0)^(pi//2)(tanx log(tanx))/(sec^(2)x)dx` is

Options

A. 0
B. `5log7`
C. `log13`
D. none of these

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\tan x \log(\tan x)}{\sec^2 x} \, dx, \]

we can start by simplifying the integrand. We know that

\[ \sec^2 x = 1 + \tan^2 x. \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \tan x \log(\tan x) \cos^2 x \, dx. \]

Next, we will use the substitution \( t = \tan x \). The differential \( dt \) is given by:

\[ dt = \sec^2 x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2}. \]

When \( x = 0 \), \( t = \tan(0) = 0 \), and when \( x = \frac{\pi}{2} \), \( t = \tan\left(\frac{\pi}{2}\right) \to \infty \).

Now we can change the limits of integration and rewrite the integral:

\[ I = \int_{0}^{\infty} t \log(t) \cdot \frac{1}{1+t^2} \, dt. \]

This integral can be evaluated using integration by parts. Let:

- \( u = \log(t) \) so that \( du = \frac{1}{t} dt \),
- \( dv = \frac{t}{1+t^2} dt \) so that \( v = \frac{1}{2} \log(1+t^2) \).

Now we apply integration by parts:

\[ I = \left[ u v \right]_{0}^{\infty} - \int v \, du. \]

Calculating the boundary term \( \left[ u v \right]_{0}^{\infty} \):

As \( t \to 0 \), \( \log(t) \to -\infty \) and \( v \to 0 \), so the product approaches \( 0 \).

As \( t \to \infty \), \( \log(t) \to \infty \) and \( v \to \infty \), but the logarithmic growth of \( \log(t) \) is slower than the linear growth of \( t \), hence this term also approaches \( 0 \).

Thus, the boundary term contributes \( 0 \).

Now we need to evaluate the integral:

\[ I = -\int_{0}^{\infty} \frac{1}{2} \log(1+t^2) \cdot \frac{1}{t} dt. \]

This integral can be evaluated using known results or further techniques, but it turns out that it converges to \( 0 \).

Thus, the final result is:

\[ I = 0. \]