If \[ I_{n}=\int_{-\pi}^{\pi} \frac{\sin n x}{\left(1+\pi^{x}\right) \sin x} d x, \quad n=0,1,2, \ldots, \] then

(A) $I_{n}=I_{n+2}$; (B) $\sum_{m=1}^{10} I_{2 m+1}=10 \pi$; (C) $\sum_{m=1}^{10} I_{2 m}=0$

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Definite Integrals

I_{n} = \int_{- π}^{π} \frac{sin n x}{( 1 + π ^{x} ) sin x} d x, n = 0, 1, 2, \dots,

then

Answer: (A), (B), (C)

Using the property

\int_{- a}^{a} f (x) d x = \int_{0}^{a} [f (x) + f (- x)] d x

and substituting

x \to - x

I_{n} = \int_{0}^{π} \frac{sin n x}{sin x} [\frac{1}{1 + π ^{x}} + \frac{π ^{x}}{1 + π ^{x}}] d x = \int_{0}^{π} \frac{sin n x}{sin x} d x

Now,

I_{n + 2} - I_{n} = \int_{0}^{π} \frac{s i n ( n + 2 ) x - s i n n x}{s i n x} d x = \int_{0}^{π} \frac{2 c o s ( n + 1 ) x s i n x}{s i n x} d x = 2 \int_{0}^{π} cos (n + 1) x d x = 0

.

Thus

I_{n} = I_{n + 2}

for all

n

.

Also,

I_{1} = \int_{0}^{π} \frac{s i n x}{s i n x} d x = π

and

I_{2} = \int_{0}^{π} 2 cos x d x = 0

.

Since

I_{n} = I_{n + 2}

, all odd-indexed integrals equal

π

and all even-indexed integrals equal

0

.

Therefore:

\sum_{m = 1}^{10} I_{2 m + 1} = 10 π

and

\sum_{m = 1}^{10} I_{2 m} = 0

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · One or more correct options