Animated Solution for Mathematics - Definite Integration: The value of $\frac{e^{-\pi /4}+{\int }_0^{\pi /4}{e}^{-x}{\tan }^{50}xdx}{{\int }_0^{\pi /4}{e}^{-x}({\tan }^{49}x+{\tan }^{51}x)dx}$ is

The expression we are evaluating is:

In the context of JEE Advanced, such expressions are designed to test your ability to identify hidden patterns. We will simplify this by breaking it down into manageable components.

Let the denominator integral be $I_1$. We focus on the integrand ${\tan }^{49}x+{\tan }^{51}x$.

Factoring out ${\tan }^{49}x$, we obtain:

Recalling the fundamental identity $1+{\tan }^{2}x={\sec }^{2}x$, the denominator integral simplifies to:

Now, consider the numerator integral $I_2={\int }_0^{\pi /4}{e}^{-x}{\tan }^{50}xdx$. To connect $I_2$ to $I_1$, we apply Integration by Parts.

Let $u={\tan }^{50}x$ and $dv=e^{-x}dx$. Consequently:

Applying the formula $\int udv=uv-\int vdu$, we get:

Evaluating the boundary terms at $x=\pi /4$ and $x=0$:

Substituting this back into our expression for $I_2$:

Since the integral term is exactly $I_1$, we have $I_2=-e^{-\pi /4}+50I_1$. Rearranging this yields:

The original expression is $\frac{e^{-\pi /4}+I_2}{I_1}$. Substituting our result:

The monster has been defeated. The final answer is 50.