JEE Main 2023 — Definite Integration Question with Solution

We're given the expression:

$$\frac{e^{-\pi /4}+{\int }_0^{\pi /4}{e}^{-x}{\tan }^{50}xdx}{{\int }_0^{\pi /4}{e}^{-x}(\tan x{)}^{49}dx+{\int }_0^{\pi /4}{e}^{-x}(\tan x{)}^{51}dx}$$

Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x}{\tan }^{n}x$ from 0 to $\pi /4$, where $n$ is an integer. Let's denote this integral as $I(n)$:

$$I(n)={\int }_0^{\pi /4}{e}^{-x}{\tan }^{n}xdx$$

We can then rewrite the original expression in terms of $I(n)$:

$$\frac{e^{-\pi /4}+I(50)}{I(49)+I(51)}$$

Now, we'll apply the method of integration by parts, which states that for two functions $u(x)$ and $v(x)$:

$$\int udv=uv-\int vdu$$

We'll choose:

$$u={\tan }^{n}x,dv=e^{-x}dx$$

Then we get:

$$du=n{\tan }^{n-1}x{\sec }^{2}xdx,v=-e^{-x}$$

Applying integration by parts, we have:

$$I(n)=-e^{-x}{\tan }^{n}x{|}_0^{\pi /4}+n{\int }_0^{\pi /4}{e}^{-x}{\tan }^{n-1}x{\sec }^{2}xdx$$

Since $\tan (\pi /4)=1$, the first term evaluates to:

$$-e^{-\pi /4}$$

The second term becomes:

$$n{\int }_0^{\pi /4}{e}^{-x}{\tan }^{n-1}x(1+{\tan }^{2}x)dx=n{\int }_0^{\pi /4}{e}^{-x}{\tan }^{n-1}xdx+n{\int }_0^{\pi /4}{e}^{-x}{\tan }^{n+1}xdx$$

This is equal to:

$$n(I(n-1)+I(n+1))$$

So we have:

$$I(n)=-e^{-\pi /4}+n(I(n-1)+I(n+1))$$

Now we can substitute $n=50$ into this equation:

$$I(50)=-e^{-\pi /4}+50(I(49)+I(51))$$

So the original expression becomes:

$$\frac{e^{-\pi /4}+I(50)}{I(49)+I(51)}=\frac{e^{-\pi /4}-e^{-\pi /4}+50(I(49)+I(51))}{I(49)+I(51)}=50$$