For $n > 0$, $\int_0^{2\pi} \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx = \dots\dots\dots$

### Analyzing the Setup The integral we are evaluating is: $$I = \int_0^{2\pi} \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx$$ The "troublemaker" in our expression is the linear factor $x$. It breaks the symmetry of the trigonometric part, making direct integration impossible. Our goal is to eliminate this $x$ using the **King's Property**, which states that: $$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$ ### The Transformation By substituting $x \to 2\pi - x$, we transform our integral into a new form. The integral becomes: $$I = \int_0^{2\pi} \frac{(2\pi - x) \sin^{2n}(2\pi - x)}{\sin^{2n}(2\pi - x) + \cos^{2n}(2\pi - x)} dx$$ Since $2n$ is an even power, we utilize the trigonometric identities $\sin^{2n}(2\pi - x) = \sin^{2n} x$ and $\cos^{2n}(2\pi - x) = \cos^{2n} x$. The denominator remains unchanged, yielding: $$I = \int_0^{2\pi} \frac{(2\pi - x) \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx$$ ### The Cancellation Now, we add our original integral $I$ and our new integral $I$. The sum $2I$ becomes: $$2I = \int_0^{2\pi} \frac{(x + 2\pi - x) \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx$$ The $x$ terms cancel out, leaving us with: $$2I = 2\pi \int_0^{2\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx$$ ### Final Calculation We can simplify the integral by reducing the limits of integration. Using the symmetry of the integrand, we can write: $$I = 2\pi \int_0^{\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx$$ Further reducing the interval to $[0, \pi/2]$, we obtain: $$I = 4\pi \int_0^{\pi/2} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} dx$$ Applying the King's Property again on the interval $[0, \pi/2]$, we get: $$I = 4\pi \int_0^{\pi/2} \frac{\cos^{2n} x}{\cos^{2n} x + \sin^{2n} x} dx$$ Adding these two forms of $I$ together, we find: $$2I = 4\pi \int_0^{\pi/2} 1 dx = 4\pi [x]_0^{\pi/2} = 2\pi^2$$ Thus, the final result is: **$I = \pi^2$**

JEE Advanced 1996

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: For $n > 0$ , $\int_{0}^{2 π} \frac{x s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x = \dots\dots\dots$

Q: Why does the power $2n$ matter so much?

The even power $2n$ is crucial because it ensures that the trigonometric functions remain positive and symmetric over the interval. Specifically, it allows the negative sign from the allied angle formula $\sin(2\pi - x) = -\sin x$ to disappear, as $(-\sin x)^{2n} = \sin^{2n} x$.

Q: How do I know when to use the King's Property?

Whenever you see a linear factor $x$ multiplying a symmetric function over a fixed interval $[a, b]$, the King's Property is almost always the intended path. It is the 'skeleton key' for integrals where the linear term prevents direct evaluation.

Visualized Solution

Define the Integral $I$

Let $I = \int_{0}^{2 π} \frac{x s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$
Identify the troublemaker term: the linear factor $x$ in the numerator.
The trigonometric part is highly symmetric over the interval $[0, 2 π]$ .

Apply King's Property

Recall the King's Property of definite integrals:
$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
For our limits $0$ to $2 π$ , we substitute $x \to 2 π - x$ .

Substitute $x \to 2 π - x$

Substitute $x \to 2 π - x$ into the integral:
$I = \int_{0}^{2 π} \frac{( 2 π - x ) s i n ^{2 n} ( 2 π - x )}{denominator} d x$

Simplify the Trigonometric Terms

Recall: $sin (2 π - x) = - sin x$
Since the power $2 n$ is even: $sin^{2 n} (2 π - x) = sin^{2 n} x$
Similarly: $cos^{2 n} (2 π - x) = cos^{2 n} x$
The integral becomes: $I = \int_{0}^{2 π} \frac{( 2 π - x ) s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$

Sum the Two Equations

Equation (1): $I = \int_{0}^{2 π} \frac{x s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$
Equation (2): $I = \int_{0}^{2 π} \frac{( 2 π - x ) s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$
Add (1) and (2): $2 I = \int_{0}^{2 π} \frac{( x + 2 π - x ) s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$

Eliminate the Linear Factor $x$

Simplify the sum: $2 I = 2 π \int_{0}^{2 π} \frac{s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$
Divide both sides by $2$ :
$I = π \int_{0}^{2 π} \frac{s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$

Reduce the Limit to $[0, π]$

Apply the property: $\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ if $f (2 a - x) = f (x)$
Here, $f (2 π - x) = f (x)$ is satisfied.
Therefore: $I = 2 π \int_{0}^{π} \frac{s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$

Reduce the Limit to $[0, \frac{π}{2}]$

Check symmetry again: $f (π - x) = f (x)$
Since $sin (π - x) = sin x$ and $cos^{2 n} (π - x) = cos^{2 n} x$
Apply the property again: $I = 4 π \int_{0}^{π /2} \frac{s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$

Apply King's Property on $[0, \frac{π}{2}]$

Apply King's Property on the interval $[0, \frac{π}{2}]$ :
Substitute $x \to \frac{π}{2} - x$
Using $sin (\frac{π}{2} - x) = cos x$ and $cos (\frac{π}{2} - x) = sin x$ :
$I = 4 π \int_{0}^{π /2} \frac{c o s ^{2 n} x}{c o s ^{2 n} x + s i n ^{2 n} x} d x$

Add and Solve

Add the two forms of $I$ on $[0, \frac{π}{2}]$ :
$2 I = 4 π \int_{0}^{π /2} \frac{s i n ^{2 n} x + c o s ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x$
$2 I = 4 π \int_{0}^{π /2} 1 \cdot d x$
$2 I = 4 π [x]_{0}^{π /2} = 4 π (\frac{π}{2}) = 2 π^{2} ⟹ I = π^{2}$

Conclusion

Key Takeaway: King's Property is highly effective for removing linear factors like $x$ .
Symmetry: Even powers allow multiple interval reductions.
Final Answer: $π^{2}$

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The integral we are evaluating is:

I = \int_{0}^{2 π} \frac{x sin ^{2 n} x}{sin ^{2 n} x + cos ^{2 n} x} d x

The "troublemaker" in our expression is the linear factor

x

. It breaks the symmetry of the trigonometric part, making direct integration impossible.

Our goal is to eliminate this

x

using the King's Property, which states that:

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

The Transformation

By substituting

x \to 2 π - x

, we transform our integral into a new form. The integral becomes:

I = \int_{0}^{2 π} \frac{( 2 π - x ) sin ^{2 n} ( 2 π - x )}{sin ^{2 n} ( 2 π - x ) + cos ^{2 n} ( 2 π - x )} d x

Since

2 n

is an even power, we utilize the trigonometric identities

sin^{2 n} (2 π - x) = sin^{2 n} x

and

cos^{2 n} (2 π - x) = cos^{2 n} x

. The denominator remains unchanged, yielding:

I = \int_{0}^{2 π} \frac{( 2 π - x ) sin ^{2 n} x}{sin ^{2 n} x + cos ^{2 n} x} d x

The Cancellation

Now, we add our original integral

I

and our new integral

I

. The sum

2 I

becomes:

2 I = \int_{0}^{2 π} \frac{( x + 2 π - x ) sin ^{2 n} x}{sin ^{2 n} x + cos ^{2 n} x} d x

The

x

terms cancel out, leaving us with:

2 I = 2 π \int_{0}^{2 π} \frac{sin ^{2 n} x}{sin ^{2 n} x + cos ^{2 n} x} d x

Final Calculation

We can simplify the integral by reducing the limits of integration. Using the symmetry of the integrand, we can write:

I = 2 π \int_{0}^{π} \frac{sin ^{2 n} x}{sin ^{2 n} x + cos ^{2 n} x} d x

Further reducing the interval to

[0, π /2]

, we obtain:

I = 4 π \int_{0}^{π /2} \frac{sin ^{2 n} x}{sin ^{2 n} x + cos ^{2 n} x} d x

Applying the King's Property again on the interval

[0, π /2]

, we get:

I = 4 π \int_{0}^{π /2} \frac{cos ^{2 n} x}{cos ^{2 n} x + sin ^{2 n} x} d x

Adding these two forms of

I

together, we find:

2 I = 4 π \int_{0}^{π /2} 1 d x = 4 π [x]_{0}^{π /2} = 2 π^{2}

Thus, the final result is:

$I = π^{2}$

Common Traps & FAQs

Wait, why can't we just use integration by parts for this?

Integration by parts would involve differentiating the trigonometric fraction, which leads to a significantly more complex expression. The King's Property is specifically designed to exploit the symmetry of the interval and the integrand, allowing the linear factor to vanish elegantly.

Why does the power

2 n

matter so much?

The even power

2 n

is crucial because it ensures that the trigonometric functions remain positive and symmetric over the interval. Specifically, it allows the negative sign from the allied angle formula

sin (2 π - x) = - sin x

to disappear, as

(- sin x)^{2 n} = sin^{2 n} x

How do I know when to use the King's Property?

Whenever you see a linear factor

x

multiplying a symmetric function over a fixed interval

[a, b]

, the King's Property is almost always the intended path. It is the 'skeleton key' for integrals where the linear term prevents direct evaluation.

Animated Solution for Mathematics - Definite Integration: For $n > 0$ , $\int_{0}^{2 π} \frac{x s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x = \dots\dots\dots$

Visualized Solution

Define the Integral $I$

Apply King's Property

Substitute $x \to 2 π - x$

Simplify the Trigonometric Terms

Sum the Two Equations

Eliminate the Linear Factor $x$

Reduce the Limit to $[0, π]$

Reduce the Limit to $[0, \frac{π}{2}]$

Apply King's Property on $[0, \frac{π}{2}]$

Add and Solve

Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Transformation

The Cancellation

Final Calculation

Similar Questions

If $I_{n} = \int_{- π}^{π} \frac{s i n n x}{( 1 + π ^{x} ) s i n x} d x, n = 0, 1, 2, ..., t h e n$

The value of $\int_{- π /2}^{π /2} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x$ is

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

The integral $\int_{0}^{π} \frac{8 x d x}{4 c o s ^{2} x + s i n ^{2} x}$ is equal to

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Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

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The integral $\int_{- 1}^{3/2} (∣ π^{2} x sin (π x) ∣) d x$ is equal to:

If $I = \int_{0}^{\frac{π}{2}} \frac{s i n ^{\frac{3}{2}} x}{s i n ^{\frac{3}{2}} x + c o s ^{\frac{3}{2}} x} d x$ , then $\int_{0}^{\frac{π}{2}} \frac{x s i n x c o s x}{s i n ^{4} x + c o s ^{4} x} d x$ equals :

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For n>0, ∫02π​sin2nx+cos2nxxsin2nx​dx=………

Visualized Solution

Define the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I

Apply King's Property

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→2π−x

Simplify the Trigonometric Terms

Sum the Two Equations

Eliminate the Linear Factor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Reduce the Limit to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,π]

Reduce the Limit to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,2π​]

Apply King's Property on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,2π​]

Add and Solve

Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Transformation

The Cancellation

Final Calculation

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Animated Solution for Mathematics - Definite Integration: For $n > 0$ , $\int_{0}^{2 π} \frac{x s i n ^{2 n} x}{s i n ^{2 n} x + c o s ^{2 n} x} d x = \dots\dots\dots$

Define the Integral $I$

Substitute $x \to 2 π - x$

Eliminate the Linear Factor $x$

Reduce the Limit to $[0, π]$

Reduce the Limit to $[0, \frac{π}{2}]$

Apply King's Property on $[0, \frac{π}{2}]$

If $I_{n} = \int_{- π}^{π} \frac{s i n n x}{( 1 + π ^{x} ) s i n x} d x, n = 0, 1, 2, ..., t h e n$

The value of $\int_{- π /2}^{π /2} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x$ is

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

The integral $\int_{0}^{π} \frac{8 x d x}{4 c o s ^{2} x + s i n ^{2} x}$ is equal to

Evaluate $\int_{0}^{π} \frac{x s i n 2 x s i n ( \frac{π}{2} c o s x )}{2 x - π} d x$

$\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ is

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

Integrate $\int_{0}^{π} \frac{e ^{c o s x}}{e ^{c o s x} + e ^{- c o s x}} d x$ .

The integral $\int_{- 1}^{3/2} (∣ π^{2} x sin (π x) ∣) d x$ is equal to:

If $I = \int_{0}^{\frac{π}{2}} \frac{s i n ^{\frac{3}{2}} x}{s i n ^{\frac{3}{2}} x + c o s ^{\frac{3}{2}} x} d x$ , then $\int_{0}^{\frac{π}{2}} \frac{x s i n x c o s x}{s i n ^{4} x + c o s ^{4} x} d x$ equals :