Let f(t) = ∫ ( (1-sin(logₑ t)) / (1-cos(logₑ t)) ) dt, t 1. If f(e^(π/2)) = -e^(π/2) and f(e^(π/4)

Q: Why do we use the substitution $u = \log_e t$? Can't we integrate it directly?

Integrating a function where the argument of the trig function is a log is extremely difficult. By substituting $u = \log_e t$, we convert the logarithmic argument into a linear one, which allows us to use standard trigonometric identities. It turns a 'nightmare' integral into a 'standard' one.

Q: How do I know which function is $f(u)$ in the form $\int e^u [f(u) + f'(u)] du$?

This is a classic JEE pattern. When you see $e^u$ multiplied by a sum of two terms, look for the term that, when differentiated, produces the other term (or part of it). In this problem, differentiating $-\cot(u/2)$ gives $\frac{1}{2}\csc^2(u/2)$, which is exactly the second term. Always test the more complex-looking term first!

Q: I got stuck calculating $\cot(\pi/8)$. Is there a trick?

Yes! Don't memorize it if you don't have to. Use the identity $\cot \theta = \frac{1 + \cos 2\theta}{\sin 2\theta}$. For $\theta = \pi/8$, $2\theta = \pi/4$. Since $\sin(\pi/4)$ and $\cos(\pi/4)$ are both $1/\sqrt{2}$, the calculation becomes very straightforward.

Question

Let $f(t) = \int \left( \frac{1-\sin(\log_e t)}{1-\cos(\log_e t)} \right) dt, t > 1$. If $f(e^{\pi/2}) = -e^{\pi/2}$ and $f(e^{\pi/4}) = \alpha e^{\pi/4}$, then $\alpha$ equals

Options:

(A) $1 + \sqrt{2}$

(B) $-1 - \sqrt{2}$

(C) $-1 - 2\sqrt{2}$

(D) $-1 + \sqrt{2}$

Sigma Percentile · Accepted Answer

### The Art of Unmasking the Integral

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of logarithms and trigonometry.

You might feel a shiver of hesitation when you see $f(t) = \int \left( \frac{1-\sin(\log_e t)}{1-\cos(\log_e t)} \right) dt$. That is perfectly normal.

In the JEE Advanced arena, the most intimidating problems are often just simple concepts wearing a scary mask. Our job is to strip away that mask.

### Phase 1: The Substitution Strategy

The presence of $\log_e t$ inside the trigonometric functions is a massive red flag. It is the 'noise' preventing us from using standard identities.

We need to simplify the domain. Let us perform the substitution $u = \log_e t$.

This implies $t = e^u$, and consequently, $dt = e^u du$. Suddenly, the integral transforms into:

$$f(t) = \int e^u \left( \frac{1-\sin u}{1-\cos u} \right) du$$

Look at that! We have moved from a logarithmic nightmare to a clean, exponential-trigonometric structure. This is the first victory.

### Phase 2: Trigonometric Surgery

Now, we must operate on the fraction $\frac{1-\sin u}{1-\cos u}$. In calculus, whenever you see $1 - \cos u$ or $1 + \cos u$, your brain should immediately trigger the half-angle identities.

We know that $1 - \cos u = 2\sin^2(u/2)$. Similarly, to maintain consistency, we expand $\sin u$ as $2\sin(u/2)\cos(u/2)$.

Substituting these into our fraction, we get:

$$\frac{1 - 2\sin(u/2)\cos(u/2)}{2\sin^2(u/2)}$$

By splitting this into two terms, we get $\frac{1}{2}\csc^2(u/2) - \cot(u/2)$. This simplifies the expression beautifully.

### Phase 3: The Golden Form

We are now looking at the integral $\int e^u [-\cot(u/2) + \frac{1}{2}\csc^2(u/2)] du$. Does this look familiar? It should!

This is the legendary $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$ pattern. If we define $f(u) = -\cot(u/2)$, then its derivative $f'(u)$ is indeed $\frac{1}{2}\csc^2(u/2)$.

The integral collapses instantly into $-e^u \cot(u/2) + C$. Replacing $u$ with $\log_e t$, we get:

$$f(t) = -t \cot\left(\frac{\log_e t}{2}\right) + C$$

### Phase 4: Finding the Constants and the Final Answer

We are given $f(e^{\pi/2}) = -e^{\pi/2}$. Substituting this into our result, we find that $C = 0$.

This simplifies our function to $f(t) = -t \cot\left(\frac{\log_e t}{2}\right)$. Finally, we evaluate $f(e^{\pi/4}) = \alpha e^{\pi/4}$.

This leads us to $\alpha = -\cot(\pi/8)$. Using the identity $\cot(\pi/8) = \sqrt{2} + 1$, we arrive at our final answer:

**$$\alpha = -1 - \sqrt{2}$$**

See? The monster was just a puzzle waiting to be solved. Keep this logic in your toolkit, and no integral will ever scare you again.

Animated Solution for Mathematics - Indefinite Integration: Let $f (t) = \int (\frac{1 - s i n ( l o g _{e} t )}{1 - c o s ( l o g _{e} t )}) d t, t > 1$ . If $f (e^{π /2}) = - e^{π /2}$ and $f (e^{π /4}) = α e^{π /4}$ , then $α$ equals

Select Answer:

Visualized Solution

Substitution $u = lo g_{e} t$

Rewriting the Integral

Simplifying the Denominator

Expanding the Numerator

Splitting the Fraction

Identifying the Special Integral Form

Verifying the Derivative

Evaluating the Integral

Applying the First Boundary Condition

Solving for the Constant $C$

Setting up the Second Condition

Evaluating $cot (\frac{π}{8})$

Finding the Final Value of $α$

The Sigma Insight: Evaluation of Special Integral Forms

The Art of Unmasking the Integral

Phase 1

Phase 2

Phase 3

Phase 4

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