Show that ∫(0 to nπ + v) |sin x| dx = 2n + 1 - cos v where n is a positive integer and 0 ≤ v

Q: Why can we just drop the absolute value bars?

Great question! The absolute value function $|\sin x|$ is always non-negative. In the interval $[0, \pi]$, $\sin x$ is non-negative, so $|\sin x| = \sin x$. Because the function is periodic with period $\pi$, this holds true for every interval $[k\pi, (k+1)\pi]$. Thus, we can safely remove the bars.

Q: Why is the integral of $|\sin x|$ from $0$ to $\pi$ equal to 2?

Let's calculate it: $\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$. This is a fundamental result. Think of it as the area of one 'hump' of the sine wave.

Q: What if $v$ is not between $0$ and $\pi$?

The problem states $0 \leq v < \pi$. If $v$ were larger, we would simply include more full humps in the first part of our split. The method remains the same: isolate the full humps and handle the remainder.

Question

Show that $\int_{0}^{n \pi+v}|\sin x| d x=2 n+1-\cos v$ where $n$ is a positive integer and $0 \leq v<\pi$.

Sigma Percentile · Accepted Answer

### The Beauty of the Sine Wave

Welcome, fellow traveler of the mathematical landscape! Today, we are going to explore a problem that beautifully bridges the gap between geometry and calculus.

We are tasked with proving that $\int_{0}^{n\pi+v} |\sin x| dx = 2n + 1 - \cos v$. At first glance, this might look like a daunting integral, but once we peel back the layers, you will see the elegance hidden within.

### Phase 1: Visualizing the Humps

Imagine the standard sine wave, $y = \sin x$. It oscillates gracefully between $-1$ and $1$.

Now, apply the absolute value function: $f(x) = |\sin x|$. The negative portions of the wave, which were dipping below the $x$-axis, are flipped upwards.

This creates a continuous, repeating pattern of humps, all sitting above the $x$-axis. Each of these humps has a width of exactly $\pi$, meaning the function is periodic with a fundamental period of $\pi$.

### Phase 2: The Power of Periodicity

We need to find the area under this curve from $x = 0$ to $x = n\pi + v$. Since $n$ is a positive integer, $n\pi$ represents the end of exactly $n$ full humps, and $v$ is a leftover piece where $0 \leq v < \pi$.

This suggests a natural split for our integral:

$$I = \int_{0}^{n\pi} |\sin x| dx + \int_{n\pi}^{n\pi+v} |\sin x| dx$$

By breaking the journey into these two parts, we simplify the problem significantly. The first integral covers $n$ complete humps, and because each hump is identical, we use the property of periodicity:

$$\int_{0}^{nT} f(x) dx = n \int_{0}^{T} f(x) dx$$

Here, $T = \pi$. So, the first part becomes $n \int_{0}^{\pi} |\sin x| dx$. Since $\sin x \geq 0$ on $[0, \pi]$, we can drop the absolute value bars: $n \int_{0}^{\pi} \sin x dx$.

### Phase 3: The Final Integration

Now, let's evaluate the first part. The integral of $\sin x$ is $-\cos x$.

Evaluating from $0$ to $\pi$ gives us:

$$[-\cos x]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$$

Thus, the first part is $2n$.

For the second part, $\int_{n\pi}^{n\pi+v} |\sin x| dx$, we use periodicity again. The shape of the curve from $n\pi$ to $n\pi+v$ is identical to the shape from $0$ to $v$.

So, we shift the limits: $\int_{0}^{v} |\sin x| dx$. Since $0 \leq v < \pi$, $\sin x$ is non-negative, yielding:

$$\int_{0}^{v} \sin x dx = [-\cos x]_{0}^{v} = -(\cos v - \cos 0) = 1 - \cos v$$

### Conclusion

Adding the two parts together, we get **$2n + 1 - \cos v$**.

It is a perfect result, isn't it? We have successfully decomposed a complex integral into simple, manageable pieces using the inherent symmetry of the sine function. Keep this geometric intuition in your toolkit—it will serve you well in many more JEE problems to come!

Animated Solution for Mathematics - Definite Integration: Show that $\int_{0}^{nπ + v} ∣ sin x ∣ d x = 2 n + 1 - cos v$ where $n$ is a positive integer and $0 \leq v < π$ .

Visualized Solution

Visualizing $f (x) = ∣ sin x ∣$

Splitting the Integral

Setting up the Split

Applying Periodicity

Simplifying the First Integral

Simplifying the Second Integral

Removing Absolute Value

Evaluating the First Integral

Evaluating the Second Integral

Final Result

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of the Sine Wave

Phase 1

Phase 2

Phase 3

Conclusion

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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; } Show that ∫0nπ+v​∣sinx∣dx=2n+1−cosv where n is a positive integer and 0≤v<π.