Evaluate the following ∫ (0 to 1) (x sin⁻¹ x) / √(1-x²) dx.

Q: Wait, why do we have to change the limits of integration? Can't we just integrate and then substitute back?

You absolutely can substitute back, but changing the limits is a much more elegant and less error-prone approach. When you change the variable from $x$ to $\theta$, the boundaries of the integral must correspond to the new variable. If you don't change them, you are essentially mixing two different coordinate systems, which is a recipe for disaster. Changing them keeps your logic consistent and clean.

Q: How do I know which function to pick as $u$ in Integration by Parts?

Great question! We use the ILATE rule, which stands for Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, and Exponential functions. In our integral $\int \theta \sin \theta d\theta$, $\theta$ is Algebraic and $\sin \theta$ is Trigonometric. Since 'A' comes before 'T' in ILATE, we choose $u = \theta$ and $dv = \sin \theta d\theta$. This choice makes the subsequent integral simpler, which is the whole point of the method!

Q: Why does $\sqrt{1-\sin^2 \theta}$ become $\cos \theta$ and not something else?

Remember the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$. This means $\cos^2 \theta = 1 - \sin^2 \theta$, so $\sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = |\cos \theta|$. Since our limits are from $0$ to $\frac{\pi}{2}$, the cosine function is positive in this interval, so we can safely write it as $\cos \theta$.

Question

Evaluate the following $\int_0^1 \frac{x \sin^{-1} x}{\sqrt{1-x^2}} dx$.

Sigma Percentile · Accepted Answer

### The Beauty of Transformation: Solving the Integral of $\frac{x \sin^{-1} x}{\sqrt{1-x^2}}$

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a problem that might seem daunting at first glance: the integral of $\frac{x \sin^{-1} x}{\sqrt{1-x^2}}$ from $0$ to $1$.

It looks like a tangled mess, doesn't it? But in the world of JEE Advanced, complexity is often just a mask for elegance. Let's peel back the layers together.

### Phase 1: The Detective Work

When you look at an integral, don't just see a collection of symbols. See the relationships. We have $x$, we have $\sin^{-1} x$, and we have $\frac{1}{\sqrt{1-x^2}}$ hiding in the denominator.

A seasoned eye immediately spots the relationship: the derivative of $\sin^{-1} x$ is exactly $\frac{1}{\sqrt{1-x^2}}$. This is our golden ticket. It tells us that the integrand is not just a random collection of functions; it is a structured, solvable puzzle.

### Phase 2: The Transformation

We are going to shift our entire problem from the $x$-world into the $	heta$-world. We set $	heta = \sin^{-1} x$, which immediately implies $x = \sin 	heta$.

Now, we need to handle the differential $dx$. Differentiating $x = \sin 	heta$ with respect to $	heta$ gives us:
$$dx = \cos 	heta \, d	heta$$

But wait! We must not forget the limits. When $x = 0$, $	heta = \sin^{-1}(0) = 0$. When $x = 1$, $	heta = \sin^{-1}(1) = \frac{\pi}{2}$. Our new boundaries are set from $0$ to $\frac{\pi}{2}$.

### Phase 3: The Simplification

Now, let's plug these values into our integral:
$$I = \int_0^{\frac{\pi}{2}} \frac{(\sin 	heta) \cdot 	heta}{\sqrt{1-\sin^2 	heta}} \cdot \cos 	heta \, d	heta$$

Look at the denominator. $\sqrt{1-\sin^2 	heta}$ is simply $\cos 	heta$. The $\cos 	heta$ in the denominator and the $\cos 	heta$ from our $dx$ term cancel out beautifully!

We are left with the much simpler integral:
$$I = \int_0^{\frac{\pi}{2}} 	heta \sin 	heta \, d	heta$$

### Phase 4: The Integration by Parts

We have arrived at a product of two functions: an algebraic function, $	heta$, and a trigonometric function, $\sin 	heta$. This is the perfect candidate for Integration by Parts. Using the ILATE rule, we choose $u = 	heta$ and $dv = \sin 	heta \, d	heta$.

Applying the formula $\int u \, dv = uv - \int v \, du$, we get:
$$I = [	heta (-\cos 	heta)]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} (1)(-\cos 	heta) \, d	heta$$

The two minus signs multiply to become a plus, giving us:
$$I = [-	heta \cos 	heta]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos 	heta \, d	heta$$

The integral of $\cos 	heta$ is $\sin 	heta$. So, our final expression is:
$$[-	heta \cos 	heta + \sin 	heta]_0^{\frac{\pi}{2}}$$

### Phase 5: The Final Evaluation

Now, we just need to plug in the limits. At the upper limit $	heta = \frac{\pi}{2}$:
$$-(\frac{\pi}{2}) \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) = -(\frac{\pi}{2})(0) + 1 = 1$$

At the lower limit $	heta = 0$:
$$-(0) \cos(0) + \sin(0) = 0 + 0 = 0$$

Subtracting the lower limit from the upper limit, we get $1 - 0 = 1$.

And there it is! The complexity vanishes, leaving behind a clean, elegant result. The final answer is **$1$**.

Animated Solution for Mathematics - Definite Integration: Evaluate the following $\int_{0}^{1} \frac{x s i n ^{- 1} x}{1 - x ^{2}} d x$ .

Enter Numerical Value:

Visualized Solution

Analyze the Integrand

Identify the Pattern

Substitution Setup

Differential Transformation

Changing the Limits

Rewriting the Integral

Simplifying the Integrand

Integration by Parts (ILATE)

Applying the Formula

Final Integration

Evaluating Limits

Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of Transformation

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate the following ∫01​1−x2​xsin−1x​dx.