Definite Integration

$f(x)=x+\int\limits_{0}^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$ $f(x)=x+\int\limits_{0}^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x)\quad....(1)$ On comparing with $f(x)=x+\dfrac{a}{\pi^{2}-4} \sin x+\dfrac{b}{\pi^{2}-4} \cos x, x \in \mathbb{R}$ then $\Rightarrow \dfrac{a}{\pi^{2}-4}=\int_{0}^{\pi / 2} \cos y f(y) d y \quad....(2)$ $\Rightarrow \dfrac{b}{\pi^{2}-4}=\int_{0}^{\pi / 2} \sin y f(y) d y \quad....(3)$ Add (2) and (3) $\dfrac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f(y) d y \quad....(4)$ $\dfrac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f\left(\dfrac{\pi}{2}-y\right) d y\quad....(5)$ Add (4) and (5) $\dfrac{2(a+b)}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y)\left(\dfrac{\pi}{2}+\dfrac{(a+b)}{\pi^{2}-4}(\sin y+\cos y)\right) d y$ $=\pi+\dfrac{a+b}{\pi^{2}-4}\left(\dfrac{\pi}{2}+1\right)$ $(a+b)=-2 \pi(\pi+2)$

We have $\int \dfrac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \dfrac{x}{a}+C$ Hence $\int_{\dfrac{3 \sqrt{2}}{4}}^{\dfrac{3 \sqrt{3}}{4}} \dfrac{48}{\sqrt{9-4 x^2}} d x=\dfrac{48}{2} \times\left[\sin ^{-1} \dfrac{2 x}{3}\right]_{\dfrac{3 \sqrt{2}}{4}}^{\dfrac{3 \sqrt{3}}{4}}$ $=24 \times\left[\sin ^{-1}\left(\dfrac{2}{3} \times \dfrac{3 \sqrt{3}}{4}\right)-\sin ^{-1}\left(\dfrac{2}{3} \times \dfrac{3 \sqrt{2}}{4}\right)\right]$ $=24 \times\left[\sin ^{-1} \dfrac{\sqrt{3}}{2}-\sin ^{-1} \dfrac{1}{\sqrt{2}}\right]$ $=24 \times\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)$ $=24 \times \dfrac{\pi}{12}=2 \pi$

The given integral is:

.............(i) Perform a substitution, $x \rightarrow 1-x$. This gives:

Simplify the expression inside the integral:

............(ii) Add the original integral (i) and the integral after substitution (ii):

Factor the quadratic expression in the denominator:

To solve this integral, we can perform a change of variables using the substitution $x-\dfrac{1}{2}= \dfrac{\sqrt{11}}{2}\tan{u}$, then $dx=\dfrac{\sqrt{11}}{2}\sec^2{u} du$:

Using the identity $\sec^2{u}=1+\tan^2{u}$:

Now we need to find the limits of the integral after the substitution.

If $x=0$, then $u=\tan^{-1}\left(-\dfrac{1}{\sqrt{11}}\right)$.

If $x=1$, then $u=\tan^{-1}\left(\dfrac{1}{\sqrt{11}}\right)$.

So, the integral becomes:

Using the properties of the arctangent function, we can rewrite the integral as:

From this result, we have $\alpha=\sqrt{11}$ and $\beta=\sqrt{10}$. Now, we can find $\alpha^4 - \beta^4$:

Thus, $\alpha^4 - \beta^4 = 21$.

We're given the expression:

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 $I(n)$:

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

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

We'll choose:

Then we get:

Applying integration by parts, we have:

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

The second term becomes:

This is equal to:

So we have:

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

So the original expression becomes:

Let $L=\lim\limits_{n \rightarrow \infty}\left\{\left(2^{\dfrac{1}{2}}-2^{\dfrac{1}{3}}\right)\left(2^{\dfrac{1}{2}}-2^{\dfrac{1}{5}}\right) \ldots\left(2^{\dfrac{1}{2}}-2^{\dfrac{1}{2 n-1}}\right)\right\}$ Here, $\left(2^{\dfrac{1}{2}}-2^{\dfrac{1}{3}}\right)^n<\left(2^{\dfrac{1}{2}}-2^{\dfrac{1}{3}}\right)\left(2^{\dfrac{1}{3}}-2^{\dfrac{1}{5}}\right)$

Since, $\lim\limits_{n \rightarrow \infty}\left(2^{\dfrac{1}{2}}-2^{\dfrac{1}{3}}\right)^n=0$ and $\lim\limits_{n \rightarrow \infty}\left(2^{\dfrac{1}{2}}-2^{\dfrac{1}{2 n-1}}\right)^n=0$

The integral equation is given by :

Step 1 : Break the first integral into two parts :

3.

Apply the King's property, $\int\limits_a^b f(x) dx = \int\limits_a^b f(a+b-x) dx$, to the first integral and substitute $x - \dfrac{\pi}{4} = t$ in the second integral.

This gives :

Then, noticing that $2 \sin \dfrac{\pi}{4} = \sqrt{2}$, you can factor out the term $\cos x$:

In order for this equation to hold true, either the integral of the function is zero, or the term outside the integral is zero.

Since we have no reason to assume that the integral of the function is zero, we set the term outside the integral to zero, yielding the solution:

So, the correct answer to the original problem is $\alpha = -\sqrt{2}$, which corresponds to Option C.

Let $e^x=t \Rightarrow e^x d x=d t$ When, $x \rightarrow-\log _e 2$, then $t \rightarrow \dfrac{1}{2}$ When, $x \rightarrow \log _e 2$, then $t \rightarrow 2$

...........(i) On applying integration by part method in Eq. (i), we get

.............(ii) Let $\quad I_1=\int_{1 / 2}^2 \dfrac{2 t}{\sqrt{1+t^2}} d t$ Let $1+t^2=w$ $2 t d t=d w$ When, $t \rightarrow \dfrac{1}{2}$, then $w=\dfrac{5}{4}$ When, $t \rightarrow 2$, then $w=5$

On substitute value of $I_1$ in Eq. (ii), we get