Let $a \in \left( 0 , \frac { \pi } { 2 } \right)$, then the value of
$\lim _ { a \rightarrow 0 } \frac { 1 } { a ^ { 3 } } \int _ { 0 } ^ { a } \ell n (1+tan a tan x)dx$ is equal to 

We are to evaluate the limit:

$\lim _{a \rightarrow 0} \frac{1}{a^{3}} \int_{0}^{a} \ln (1+\tan a \tan x) d x$

Step 1: Approximate for small $\boldsymbol{a}$

Since $a \rightarrow 0$, we can use the small-angle approximations:

- $\tan a \approx a+\frac{a^{3}}{3}$

- $\tan x \approx x+\frac{x^{3}}{3}$

Thus, the integrand becomes:

$\ln (1+\tan a \tan x) \approx \ln \left(1+\left(a+\frac{a^{3}}{3}\right)\left(x+\frac{x^{3}}{3}\right)\right)$

For small $a$ and $x$, the higher-order terms can be neglected, so:

$\ln (1+\tan a \tan x) \approx \ln (1+a x)$

Step 2: Expand $\ln (1+a x)$ using Taylor series

The Taylor series expansion of $\ln (1+y)$ around $y=0$ is:

$\ln (1+y)=y-\frac{y^{2}}{2}+\frac{y^{3}}{3}-\cdots$

Substituting $y=a x$ :

$\ln (1+a x)=a x-\frac{(a x)^{2}}{2}+\frac{(a x)^{3}}{3}-\cdots$

Step 3: Integrate term by term

Now, integrate $\ln (1+a x)$ from 0 to $a$ :

$\int_{0}^{a} \ln (1+a x) d x \approx \int_{0}^{a}\left(a x-\frac{a^{2} x^{2}}{2}+\frac{a^{3} x^{3}}{3}\right) d x$

Compute each term separately:

$\begin{array}{c} \int_{0}^{a} a x d x=a \cdot \frac{a^{2}}{2}=\frac{a^{3}}{2} \\ \int_{0}^{a} \frac{a^{2} x^{2}}{2} d x=\frac{a^{2}}{2} \cdot \frac{a^{3}}{3}=\frac{a^{5}}{6} \\ \int_{0}^{a} \frac{a^{3} x^{3}}{3} d x=\frac{a^{3}}{3} \cdot \frac{a^{4}}{4}=\frac{a^{7}}{12} \end{array}$

Thus, the integral becomes:

$\int_{0}^{a} \ln (1+a x) d x \approx \frac{a^{3}}{2}-\frac{a^{5}}{6}+\frac{a^{7}}{12}$

Step 4: Compute the limit

Now, divide by $a^{3}$ and take the limit as $a \rightarrow 0$ :

$\lim _{a \rightarrow 0} \frac{1}{a^{3}}\left(\frac{a^{3}}{2}-\frac{a^{5}}{6}+\frac{a^{7}}{12}\right)=\lim _{a \rightarrow 0}\left(\frac{1}{2}-\frac{a^{2}}{6}+\frac{a^{4}}{12}\right)$

As $\boldsymbol{a} \rightarrow \mathbf{0}$, the higher-order terms vanish, and we are left with:

$\lim _{a \rightarrow 0} \frac{1}{a^{3}} \int_{0}^{a} \ln (1+\tan a \tan x) d x=\frac{1}{2}$

Final Answer

$\frac{1}{2}$