If ∫_{sin x}^1 t² f(t) dt = 1 - sin x, then f(1/√3) is (a) 1/3 (b) 1/√3 (c) 3 (d) √3

Q: Wait, why can't we just integrate $f(t)$ directly?

Because $f(t)$ is an unknown function. We don't know its form, so we can't integrate it. Differentiation is the only way to 'undo' the integral and isolate $f$. Think of it as peeling back a layer of an onion to see what is inside.

Q: Do I really need the chain rule when differentiating the lower limit?

Absolutely! The Leibniz rule is essentially the Fundamental Theorem of Calculus combined with the Chain Rule. If you forget to multiply by $\frac{d}{dx}(\sin x)$, you will miss the $\cos x$ term, and the whole equation will not balance. Always respect the chain rule!

Q: What if $\cos x = 0$?

In the context of this problem, we are looking for a specific value of $f$. As long as we are not at a point where the division is undefined, the algebraic relationship holds. We are essentially finding the functional form of $f$ that satisfies the equation for all valid $x$.

Question

If $\int_{\sin x}^1 t^2 f(t) dt = 1 - \sin x$, then $f\left(\frac{1}{\sqrt{3}}\right)$ is

Options:

(A) $1/3$

(B) $1/\sqrt{3}$

(C) $3$

(D) $\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a locked vault. The vault is an integral, and inside, hidden away, is the function $f(t)$.

You are given the equation:

$$
\int_{\sin x}^1 t^2 f(t) dt = 1 - \sin x
$$

Your mission is to extract $f(t)$ and find its value at a specific point. This is not just a calculation; it is a detective story.

### The Newton-Leibniz Key

To break into this vault, we need the right tool. We cannot simply integrate $f(t)$ because we do not know what it is.

Instead, we must use the **Newton-Leibniz rule**, also known as the Leibniz Integral Rule. This rule is the master key for differentiating integrals where the limits are functions of $x$.

It states that:

$$
\frac{d}{dx} \int_{u(x)}^{v(x)} g(t) dt = g(v(x)) \cdot v'(x) - g(u(x)) \cdot u'(x)
$$

### The Differentiation Attack

Let us apply the operator $\frac{d}{dx}$ to both sides of our equation. On the right side, the derivative of $1 - \sin x$ is straightforward: it is $-\cos x$.

On the left side, we apply the Newton-Leibniz rule. The upper limit is $1$, a constant, so its derivative is $0$. This is a stroke of luck, as the first term of the rule, $g(1) \cdot 0$, vanishes completely.

Now for the lower limit: we substitute $t = \sin x$ into our integrand $t^2 f(t)$, giving us $\sin^2 x \cdot f(\sin x)$. Crucially, we must multiply by the derivative of the lower limit, which is $\frac{d}{dx}(\sin x) = \cos x$.

Putting it all together, the left side becomes:

$$
0 - (\sin^2 x \cdot f(\sin x)) \cdot \cos x
$$

### The Algebraic Reveal

Now we equate our results:

$$
-\sin^2 x \cdot f(\sin x) \cdot \cos x = -\cos x
$$

The negative signs cancel out, and the $\cos x$ terms appear on both sides. Assuming $\cos x 
eq 0$, we can divide both sides by $\cos x$ to simplify our expression to:

$$
\sin^2 x \cdot f(\sin x) = 1
$$

This reveals the hidden function:

$$
f(\sin x) = \frac{1}{\sin^2 x}
$$

### The Final Calculation

The problem asks for $f\left(\frac{1}{\sqrt{3}}ight)$. This means we need to evaluate our function when the input is $\frac{1}{\sqrt{3}}$.

Since our function is defined as $f(\sin x) = \frac{1}{\sin^2 x}$, we simply set $\sin x = \frac{1}{\sqrt{3}}$. Substituting this into our expression, we get:

$$
f\left(\frac{1}{\sqrt{3}}ight) = \frac{1}{\left(\frac{1}{\sqrt{3}}ight)^2}
$$

Squaring $\frac{1}{\sqrt{3}}$ gives us $\frac{1}{3}$. Therefore:

$$
f\left(\frac{1}{\sqrt{3}}ight) = \frac{1}{1/3} = 3
$$

The mystery is solved, and the final answer is **3**. Remember, in JEE Advanced, the most complex-looking integrals are often just waiting for the right rule to unlock them.

Animated Solution for Mathematics - Definite Integration: If $\int_{s i n x}^{1} t^{2} f (t) d t = 1 - sin x$ , then $f (\frac{1}{3})$ is

Select Answer:

Visualized Solution

The Integral Equation

Newton-Leibniz Rule

Differentiating Both Sides

LHS: Upper Limit

LHS: Lower Limit

LHS: Final Expression

Differentiating the RHS

Equating LHS and RHS

Isolating $f (sin x)$

Finding $f (\frac{1}{3})$

Final Calculation

Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Newton-Leibniz Key

The Differentiation Attack

The Algebraic Reveal

The Final Calculation

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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; } If ∫sinx1​t2f(t)dt=1−sinx, then f(3​1​) is