The value of the integral ∫_{0}^{π/2} 60 sin(6x)/sin x dx is equal to

Q: Wait, why can't I just use the expansion of $\sin(6x)$? That seems like a nightmare.

You are absolutely right! Expanding $\sin(6x)$ using multiple-angle formulas would lead to a massive polynomial in $\sin x$ and $\cos x$, which would be an algebraic trap. The beauty of the telescoping identity is that it bypasses the expansion entirely, turning a complex ratio into a simple sum of cosines.

Q: How do I know when to stop the recursion?

The recursion stops when the numerator becomes $\sin(2x)$ or $\sin(0x)$. Since $\frac{\sin(2x)}{\sin x} = 2\cos x$, you have reached a point where every term is a simple cosine function, which is exactly what we need for integration.

Q: Is there a trick to remembering the identity $\sin(nx) - \sin((n-2)x) = 2\cos((n-1)x)\sin x$?

Think of it as the 'difference of sines' formula. If you have $\sin(A) - \sin(B) = 2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$, let $A = nx$ and $B = (n-2)x$. Then $\frac{A+B}{2} = (n-1)x$ and $\frac{A-B}{2} = x$. It is just a standard identity in disguise!

Q: Why does the lower limit $0$ result in $0$ for all terms?

Because the integral of $\cos(kx)$ is $\frac{\sin(kx)}{k}$. At the lower limit $x=0$, we have $\sin(0)$, which is always $0$. It is a gift from the math gods that simplifies our calculation significantly!

Q: I got $104$. Is it possible to get a negative value for this integral?

Great question! Since $\sin(6x)$ and $\sin x$ are both positive in the interval $(0, \pi/6)$ and oscillate thereafter, we must be careful. However, the integral of this specific ratio over $(0, \pi/2)$ is strictly positive. If you get a negative result, check your evaluation of $\sin(3\pi/2)$—it is $-1$, not $1$!

Question

The value of the integral $\int_{0}^{\pi/2} 60 \frac{\sin(6x)}{\sin x} dx$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we encounter an integral that, at first glance, looks like a formidable wall:

$$I = \int_{0}^{\pi/2} 60 \frac{\sin(6x)}{\sin x} dx$$

Many students see the ratio $\frac{\sin(6x)}{\sin x}$ and immediately reach for the multiple-angle expansion formulas. I urge you: stop. If you expand $\sin(6x)$, you will find yourself drowning in a sea of powers of $\sin x$ and $\cos x$. Instead, let us look for the hidden symmetry.

### The Telescoping Secret

Imagine you are a detective looking for a pattern. We have the identity $\sin(nx) - \sin((n-2)x) = 2\cos((n-1)x)\sin x$. This is our skeleton key.

By rearranging this, we get the recursive relation:

$$\frac{\sin(nx)}{\sin x} = \frac{\sin((n-2)x)}{\sin x} + 2\cos((n-1)x)$$

This is not just algebra; it is a decomposition. We are breaking a complex frequency into a sum of simpler, harmonic components.

Let us apply this for $n=6, 4, 2$. For $n=6$, we get $\frac{\sin(6x)}{\sin x} = \frac{\sin(4x)}{\sin x} + 2\cos(5x)$.

Then, for $n=4$, we peel back another layer: $\frac{\sin(4x)}{\sin x} = \frac{\sin(2x)}{\sin x} + 2\cos(3x)$. Finally, for $n=2$, we reach the base: $\frac{\sin(2x)}{\sin x} = 2\cos x$.

When we combine these, the intermediate terms vanish like magic, leaving us with a beautiful, elegant sum:

$$\frac{\sin(6x)}{\sin x} = 2\cos(5x) + 2\cos(3x) + 2\cos(x)$$

### The Integration Phase

Now, the monster has been tamed. Our integral transforms into something we can handle with ease:

$$I = 120 \int_{0}^{\pi/2} (\cos(5x) + \cos(3x) + \cos(x)) dx$$

Integrating term by term, we apply the fundamental theorem of calculus. The integral of $\cos(ax)$ is simply $\frac{\sin(ax)}{a}$.

Thus, we have:

$$I = 120 \left[ \frac{\sin(5x)}{5} + \frac{\sin(3x)}{3} + \sin(x) ight]_0^{\pi/2}$$

### The Final Evaluation

As we substitute the upper limit $x = \frac{\pi}{2}$, we must be precise. We know that $\sin(\frac{5\pi}{2}) = 1$, $\sin(\frac{3\pi}{2}) = -1$, and $\sin(\frac{\pi}{2}) = 1$.

Substituting these values, the expression inside the bracket becomes:

$$\left( \frac{1}{5} - \frac{1}{3} + 1 ight)$$

When we evaluate at the lower limit $x=0$, every term vanishes because $\sin(0) = 0$. We are left with the arithmetic of fractions:

$$\frac{1}{5} - \frac{1}{3} + 1 = \frac{3 - 5 + 15}{15} = \frac{13}{15}$$

Finally, we multiply by our constant $120$:

$$I = 120 	imes \frac{13}{15} = 8 	imes 13 = 104$$

There it is. The complexity has dissolved, leaving behind a clean, integer result of **104**. Remember, in JEE Advanced, the most difficult-looking problems are often hiding a simple, elegant structure.

Animated Solution for Mathematics - Definite Integration: The value of the integral $\int_{0}^{π /2} 60 \frac{s i n ( 6 x )}{s i n x} d x$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Integral

The Telescoping Identity

Decomposing $\frac{s i n ( 6 x )}{s i n x}$ - Part 1

Decomposing $\frac{s i n ( 6 x )}{s i n x}$ - Part 2

Combining the Terms

Substitute into Integral

Integrate Term by Term

Substitute Upper Limit

Evaluate Trig Values

Substitute Lower Limit

Simplify Fractions

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Telescoping Secret

The Integration Phase

The Final Evaluation

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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; } The value of the integral ∫0π/2​60sinxsin(6x)​dx is equal to