Q109 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics

$\int e^{4 x^2+8 x-4}(x+1) \cos \left(3 x^2+6 x-4\right) d x=$

  • A
    $\frac{e^{4 x^2+8 x-4}}{25}\left[3 \sin \left(3 x^2+6 x-4\right)-4 \cos \left(3 x^2+6 x-4\right)\right]+c$
  • Correct answer
    B
    $\frac{e^{4 x^2+8 x-4}}{50}\left[4 \cos \left(3 x^2+6 x-4\right)+3 \sin \left(3 x^2+6 x-4\right)\right]+c$
  • C
    $\frac{e^{4 x^2+8 x-4}}{25}\left[3 \cos \left(3 x^2+6 x-4\right)+4 \sin \left(3 x^2+6 x-4\right)\right]+c$
  • D
    $\frac{e^{4 x^2+8 x-4}}{50}\left[4 \sin \left(3 x^2+6 x-4\right)-3 \cos \left(3 x^2+6 x-4\right)\right]+c$

Solution

$I=\int e^{4 x^2+8 x-4}(x+1) \cos \left(3 x^2+6 x-4\right) d x$
Let $x^2+2 x=t \Rightarrow(x+1) d x=\frac{d t}{2}$ $I=\frac{1}{2} \int e^{4 t-4} \cos (3 t-4) d t$
We know, $\int e^{a x} \cos (b x+c) d x$ $=\frac{e^{a x}}{a^2+b^2}(a \cos (b x+c)+b \sin (b x+c))$ $\begin{aligned} & \Rightarrow I=\frac{1}{2} \frac{e^{4 t-4}}{16+9}(4 \cos (3 t-4)+3 \sin (3 t-4))+c \\ & \Rightarrow I=\frac{e^{4 x^2+8 x-2}}{50}\left[4 \cos \left(3 x^2+6 x-4\right)+3 \sin \left(3 x^2+6 x-4\right)\right]+c\end{aligned}$

Hence, the correct answer is option: 'B'.

Relevant Questions