Q113 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics
If $\int \mathrm{e}^x\left(\sin ^2 2 \mathrm{x}-8 \cos 4 \mathrm{x}\right) \mathrm{dx}=\mathrm{e}^x f(\mathrm{x})+\mathrm{c}$, then $f\left(\frac{\pi}{4}\right)=$
- A$0$
- Correct answerB$1$
- C$-1$
- D$e$
Solution
$\begin{aligned} & \int e^x\left(\sin ^2 2 x-8 \cos 4 x\right) d x \\ & =\int e^x\left(\frac{1-\cos 4 x}{2}-8 \cos 4 x\right) d x \\ & =\int e^x\left(\frac{1-17 \cos 4 x}{2}\right) d x \\ & =\frac{1}{2}\left[\int e^x d x-17 \int e^x \cos 4 x d x\right] \\ & \because \int e^{a x} \cos (b x+c) d x=\frac{e^{a x}}{a^2+b^2}[a \cos (b x+c)+b \sin (b x+c)] \\ & \therefore \quad \int e^x \cos 4 x d x=\frac{e^x}{1^2+4^2}[\cos 4 x+4 \sin 4 x] \\ & \therefore \quad \frac{1}{2}\left[e^x-17 \cdot \frac{e^x}{17}(\cos 4 x+4 \sin 4 x)\right]+C \\ & \quad=e^x\left[\frac{1}{2}-\cos 4 x-4 \sin 4 x\right]+C\end{aligned}$
$\begin{aligned} \therefore \quad f(x) & =\frac{1}{2}-\cos 4 x-4 \sin 4 x \\ f\left(\frac{\pi}{4}\right) & =\frac{1}{2}-\frac{1}{2} \cos \pi-\frac{4}{2} \sin \pi \\ & =\frac{1}{2}-\frac{1}{2}(-1)-0=\frac{1}{2}+\frac{1}{2}=1 .\end{aligned}$
Hence, the correct answer is option: 'B'.