JEE Challenger
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MathMedAdv 2009 • Paper 2

Properties of Definite Integral with Exponential and Sine Terms

If
In=∫−ππsin⁡nx(1+πx)sin⁡x dx,n=0,1,2,…,I_n = \int_{-\pi}^{\pi} \frac{\sin nx}{(1+\pi^x)\sin x}\, dx, \quad n = 0, 1, 2, \dots,
then

Options

A

In=In+2I_n = I_{n+2}

Correct
B

∑m=110I2m+1=10π\sum_{m=1}^{10} I_{2m+1} = 10\pi

Correct
C

∑m=110I2m=0\sum_{m=1}^{10} I_{2m} = 0

Correct
D

In=In+1I_n = I_{n+1}

Topics & Concepts

IntegralsDefinite IntegrationProperties of Definite Integrals

Step-by-Step Solution

To find the correct options, we analyze the given integral:
In=∫−ππsin⁡nx(1+πx)sin⁡x dx,n=0,1,2,…I_n = \int_{-\pi}^{\pi} \frac{\sin nx}{(1+\pi^x)\sin x}\, dx, \quad n = 0, 1, 2, \dots

Using the definite integral property ∫−aaf(x) dx=∫0a[f(x)+f(−x)] dx\int_{-a}^{a} f(x)\, dx = \int_0^a [f(x) + f(-x)]\, dx, we define:
f(x)=sin⁡nx(1+πx)sin⁡xf(x) = \frac{\sin nx}{(1+\pi^x)\sin x}

Evaluating f(−x)f(-x):
f(−x)=sin⁡(−nx)(1+π−x)sin⁡(−x)=−sin⁡nx(1+1πx)(−sin⁡x)=πxsin⁡nx(1+πx)sin⁡xf(-x) = \frac{\sin(-nx)}{(1+\pi^{-x})\sin(-x)} = \frac{-\sin nx}{\left(1 + \frac{1}{\pi^x}\right)(-\sin x)} = \frac{\pi^x \sin nx}{(1+\pi^x)\sin x}

Adding f(x)f(x) and f(−x)f(-x):
f(x)+f(−x)=sin⁡nx(1+πx)sin⁡x+πxsin⁡nx(1+πx)sin⁡x=sin⁡nxsin⁡x(1+πx1+πx)=sin⁡nxsin⁡xf(x) + f(-x) = \frac{\sin nx}{(1+\pi^x)\sin x} + \frac{\pi^x \sin nx}{(1+\pi^x)\sin x} = \frac{\sin nx}{\sin x} \left(\frac{1+\pi^x}{1+\pi^x}\right) = \frac{\sin nx}{\sin x}

Therefore, the integral simplifies to:
In=∫0πsin⁡nxsin⁡x dxI_n = \int_0^{\pi} \frac{\sin nx}{\sin x}\, dx

Checking Option (A):

Consider the difference In+2−InI_{n+2} - I_n:
In+2−In=∫0πsin⁡(n+2)x−sin⁡nxsin⁡x dxI_{n+2} - I_n = \int_0^{\pi} \frac{\sin(n+2)x - \sin nx}{\sin x}\, dx

Using the trigonometric identity sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right):
sin⁡(n+2)x−sin⁡nx=2cos⁡((n+1)x)sin⁡x\sin(n+2)x - \sin nx = 2\cos((n+1)x)\sin x

Substituting this back into the integral:
In+2−In=∫0π2cos⁡((n+1)x)sin⁡xsin⁡x dx=2∫0πcos⁡((n+1)x) dxI_{n+2} - I_n = \int_0^{\pi} \frac{2\cos((n+1)x)\sin x}{\sin x}\, dx = 2\int_0^{\pi} \cos((n+1)x)\, dx

Since n≥0n \ge 0, n+1n+1 is a positive integer:
∫0πcos⁡((n+1)x) dx=[sin⁡((n+1)x)n+1]0π=0\int_0^{\pi} \cos((n+1)x)\, dx = \left[ \frac{\sin((n+1)x)}{n+1} \right]_0^{\pi} = 0

Thus:
In+2=InI_{n+2} = I_n
Hence, Option (A) is correct.

Checking Options (B) and (C):

Since In+2=InI_{n+2} = I_n for all n≥0n \ge 0:

  • All even indices have the same value: I2m=I0I_{2m} = I_0 for all m≥1m \ge 1.
  • All odd indices have the same value: I2m+1=I1I_{2m+1} = I_1 for all m≥0m \ge 0.

Computing I0I_0:
I0=∫0πsin⁡0sin⁡x dx=0I_0 = \int_0^{\pi} \frac{\sin 0}{\sin x}\, dx = 0
Hence, I2m=0I_{2m} = 0 for all m≥1m \ge 1. Therefore:
∑m=110I2m=∑m=1100=0\sum_{m=1}^{10} I_{2m} = \sum_{m=1}^{10} 0 = 0
Thus, Option (C) is correct.

Computing I1I_1:
I1=∫0πsin⁡xsin⁡x dx=∫0π1 dx=πI_1 = \int_0^{\pi} \frac{\sin x}{\sin x}\, dx = \int_0^{\pi} 1\, dx = \pi
Hence, I2m+1=πI_{2m+1} = \pi for all m≥0m \ge 0. Therefore:
∑m=110I2m+1=∑m=110π=10π\sum_{m=1}^{10} I_{2m+1} = \sum_{m=1}^{10} \pi = 10\pi
Thus, Option (B) is correct.

Checking Option (D):

Since I0=0I_0 = 0 and I1=πI_1 = \pi, we have I0≠I1I_0 \neq I_1, meaning In≠In+1I_n \neq I_{n+1} in general.
Thus, Option (D) is incorrect.

Conclusion:

The correct options are A, B, and C.

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