Q117 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
$\int_{-\pi / 8}^{\pi / 8} \frac{\sin ^4(4 x)}{1+e^{4 x}} d x=$
- A$\frac{3 \pi}{128}$
- B$\frac{3 \pi}{256}$
- Correct answerC$\frac{3 \pi}{64}$
- D$\frac{3 \pi}{32}$
Solution
$I=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(4 x)}{1+e^{4 x}} d x$....(i)
Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$ $\Rightarrow \quad I=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(-4 x)}{1+e^{-4 x}} d x=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{e^{4 x} \sin ^4(4 x)}{1+e^{4 x}} d x$....(ii)
Adding (i) and (ii) $\begin{aligned} & \Rightarrow 2 I=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\left(1+e^{4 x}\right) \sin ^4(4 x)}{\left(1+e^{4 x}\right)} d x \\ & \Rightarrow 2 I=\int_{\frac{\pi}{8}}^{\frac{\pi}{8}} \sin ^4(4 x) d x \Rightarrow 2 I=2 \int_0^{\frac{\pi}{8}} \sin ^4(4 x) d x \end{aligned}$
Let $4 x=t$ $\begin{aligned} & \Rightarrow I=\frac{1}{4} \int_0^{\frac{\pi}{2}} \sin ^4 t d t \Rightarrow I=\frac{1}{4} \times \frac{3 \times 1}{4 \times 2} \times \frac{\pi}{2} \\ & \Rightarrow I=\frac{3 \pi}{64} \end{aligned}$
Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$ $\Rightarrow \quad I=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(-4 x)}{1+e^{-4 x}} d x=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{e^{4 x} \sin ^4(4 x)}{1+e^{4 x}} d x$....(ii)
Adding (i) and (ii) $\begin{aligned} & \Rightarrow 2 I=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\left(1+e^{4 x}\right) \sin ^4(4 x)}{\left(1+e^{4 x}\right)} d x \\ & \Rightarrow 2 I=\int_{\frac{\pi}{8}}^{\frac{\pi}{8}} \sin ^4(4 x) d x \Rightarrow 2 I=2 \int_0^{\frac{\pi}{8}} \sin ^4(4 x) d x \end{aligned}$
Let $4 x=t$ $\begin{aligned} & \Rightarrow I=\frac{1}{4} \int_0^{\frac{\pi}{2}} \sin ^4 t d t \Rightarrow I=\frac{1}{4} \times \frac{3 \times 1}{4 \times 2} \times \frac{\pi}{2} \\ & \Rightarrow I=\frac{3 \pi}{64} \end{aligned}$
Hence, the correct answer is option: 'C'.