For x ∈ (-π/2, π/2), if y(x) = ∫ (csc x + sin x) / (csc x sec x + tan x sin² x) dx and lim_{x→(π/2)⁻

Q: Wait, why do we divide by $t^2$ in the integral $\int \frac{1 + t^2}{1 + t^4} dt$? Is there a rule for this?

That is a fantastic question! This is a classic technique used for integrals of the form $\int \frac{ax^2 + b}{cx^4 + dx^2 + e} dx$. When you see a quartic polynomial in the denominator with only even powers, dividing by $x^2$ (or $t^2$ in our case) is the standard 'key' to unlock the problem. It allows us to express the denominator as a perfect square plus a constant, which perfectly matches the derivative of the new variable $u = t - 1/t$ in the numerator. It is not just a trick; it is a structural necessity!

Q: Why did we choose $u = t - 1/t$ instead of $u = t + 1/t$?

Great observation! We choose $u = t - 1/t$ because its derivative is $du = (1 + 1/t^2) dt$, which exactly matches our numerator. If we had chosen $u = t + 1/t$, the derivative would be $du = (1 - 1/t^2) dt$, which would not match our numerator. Always look at the numerator first—it dictates the substitution!

Q: How do we know the constant of integration $C$ is zero?

We use the boundary condition provided: $\lim_{x \to (\frac{\pi}{2})^-} y(x) = 0$. By substituting $x = \frac{\pi}{2}$ into our derived expression for $y(x)$, we find that the term involving the inverse tangent becomes zero. Since the limit is given as zero, the only way for the equation to hold is if $C = 0$. Never forget to use your boundary conditions to 'fix' the constant!

Question

For $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, if $y(x) = \int \frac{\csc x + \sin x}{\csc x \sec x + 	an x \sin^2 x} dx$ and $\lim_{x 	o (\frac{\pi}{2})^-} y(x) = 0$ then $y(\frac{\pi}{4})$ is equal to

Options:

(A) $	an^{-1}(\frac{1}{\sqrt{2}})$

(B) $\frac{1}{2} 	an^{-1}(\frac{1}{\sqrt{2}})$

(C) $-\frac{1}{\sqrt{2}} 	an^{-1}(\frac{1}{2})$

(D) $\frac{1}{\sqrt{2}} 	an^{-1}(-\frac{1}{\sqrt{2}})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the arena. Today, we face a problem that looks like a tangled mess of trigonometric functions. You see $\csc x$, $\sec x$, $	an x$, and $\sin x$ all fighting for space in a single fraction.

It is designed to intimidate. But remember, in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

### The Art of Simplification

We begin with our integrand:
$$y(x) = \int \frac{\csc x + \sin x}{\csc x \sec x + 	an x \sin^2 x} dx$$

The first rule of combat is to simplify your terrain. Let us convert everything into the fundamental language of sine and cosine. The numerator, $\csc x + \sin x$, becomes $\frac{1}{\sin x} + \sin x$, which simplifies beautifully to $\frac{1 + \sin^2 x}{\sin x}$.

Now, look at the denominator: $\csc x \sec x + 	an x \sin^2 x$. This is $\frac{1}{\sin x \cos x} + \frac{\sin x}{\cos x} \cdot \sin^2 x$, which is $\frac{1}{\sin x \cos x} + \frac{\sin^3 x}{\cos x}$.

By taking a common denominator of $\sin x \cos x$, we get $\frac{1 + \sin^4 x}{\sin x \cos x}$. When we divide the numerator by the denominator, the $\sin x$ terms vanish, and the $\cos x$ flips to the top. We are left with the much cleaner integral:
$$y(x) = \int \frac{(1 + \sin^2 x) \cos x}{1 + \sin^4 x} dx$$

### The Power of Substitution

Now, look at that $\cos x dx$ sitting there. It is a beacon of hope, as it is the derivative of $\sin x$. This is our cue to perform the substitution $t = \sin x$, which gives us $dt = \cos x dx$.

The integral transforms into:
$$y = \int \frac{1 + t^2}{1 + t^4} dt$$

This is a classic form that every JEE aspirant must recognize. To solve it, we divide the numerator and denominator by $t^2$, yielding:
$$\int \frac{1 + 1/t^2}{t^2 + 1/t^2} dt$$

### The Final Transformation

We are almost there. We need a substitution for the denominator. Let $u = t - 1/t$. Then $du = (1 + 1/t^2) dt$.

The denominator $t^2 + 1/t^2$ can be rewritten as $(t - 1/t)^2 + 2$, which is $u^2 + 2$. Our integral is now the standard form:
$$\int \frac{du}{u^2 + 2}$$

Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$, we get:
$$y = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{u}{\sqrt{2}}ight) + C$$

Substituting back $u = t - 1/t$ and $t = \sin x$, we find:
$$y(x) = \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{\sin x - 1/\sin x}{\sqrt{2}} ight) + C$$

### The Final Act

We are given that $\lim_{x 	o (\frac{\pi}{2})^-} y(x) = 0$. As $x 	o \frac{\pi}{2}$, $\sin x 	o 1$, so the argument of the $	an^{-1}$ becomes zero, forcing **$C = 0$**.

Finally, we evaluate at $x = \frac{\pi}{4}$. Since $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, the argument becomes:
$$\frac{1/\sqrt{2} - \sqrt{2}}{\sqrt{2}} = -\frac{1}{2}$$

Thus, the final result is:
$$y\left(\frac{\pi}{4}ight) = -\frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{1}{2}ight)$$

You have conquered the monster!

Animated Solution for Mathematics - Indefinite Integration: For $x \in (- \frac{π}{2}, \frac{π}{2})$ , if $y (x) = \int \frac{c s c x + s i n x}{c s c x s e c x + t a n x s i n ^{2} x} d x$ and $lim_{x \to (\frac{π}{2})^{-}} y (x) = 0$ then $y (\frac{π}{4})$ is equal to

Select Answer:

Visualized Solution

Analyze the Given Integral

Simplify the Numerator

Simplify the Denominator

Combine and Simplify the Integrand

Substitution: $t = sin x$

Divide by $t^{2}$

Substitution: $u = t - \frac{1}{t}$

Integrate

Back-Substitute

Find the Constant $C$

Calculate $y (\frac{π}{4})$

Final Result

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Art of Simplification

The Power of Substitution

The Final Transformation

The Final Act

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Select Answer:

Visualized Solution

Analyze the Given Integral

Simplify the Numerator

Simplify the Denominator

Combine and Simplify the Integrand

Substitution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=sinx

Divide by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2

Substitution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u=t−t1​

Integrate

Back-Substitute

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Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(4π​)