Q28 / 57 · JEE Advanced 2009 (Paper 2) · Mathematics

If $I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x ; n=0,1,2$, $\ldots$, then

  • Correct answer
    A
    $I_n=I_{n+2}$
  • Correct answer
    B
    $\sum_{m=1}^{10} I_{2 m+1}=10 \pi$
  • Correct answer
    C
    $\sum_{m=1}^{10} I_{2 m}=0$
  • D
    $I_n=I_{n+1}$

Solution

(a) $$ \begin{aligned} & I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x \\ & I_n=\int_{-\pi}^\pi \frac{\pi^x \sin n x}{\left(1+\pi^x\right) \sin x} d x \\ & {\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]} \\ & 2 I_n=\int_{-\pi}^\pi \frac{\sin n x}{\sin x} d x \end{aligned} $$ $$ \begin{aligned} & \Rightarrow \quad 2 I_n=2 \int_0^\pi \frac{\sin n x}{\sin x} d x \\ & \Rightarrow \quad I_n=\int_0^\pi \frac{\sin n x}{\sin x} d x \\ & \text { Now, } \quad I_{n+2}-I_n \\ & =\int_0^\pi \frac{\sin (n+2) x-\sin n x}{\sin x} d x \\ & =\int_0^\pi \frac{2 \cos (n+1) x \sin x}{\sin x} d x \\ & =2\left[\frac{\sin (n+1) x}{(n+1)}\right]_0^\pi=0 \\ & \Rightarrow \quad I_{n+2}=I_n \\ & \text { (b) As } I_3=I_5=\ldots=I_{21} \\ & \therefore \sum_{m=1}^{10} I_{2 m+1}=10 I_3=10 \int_0^\pi \frac{\sin 3 x}{\sin x} d x \\ & =10 \int_0^\pi\left(3-4 \sin ^2 x\right) d x \\ & =10[3 x-2 x+\sin 2 x]_0^\pi=10 \pi \\ & \text { (c) As } \quad I_2=I_4=\ldots=I_{20} \\ & \therefore \quad \sum_{m=1}^{10} I_{2 m}=10 \int_0^\pi \frac{\sin 2 x}{\sin x} d x \\ & =20[\sin x]_0^\pi=0 \\ & \end{aligned} $$

Hence, the correct answer is options: 'A', 'B', 'C'.

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