AP EAMCET · Maths · Indefinite Integration
\(\int e^{4 x^2+8 x-4}(x+1) \cos \left(3 x^2+6 x-4\right) d x=\)
- A \(\frac{e^{4 x^2+8 x-4}}{25}\left[3 \sin \left(3 x^2+6 x-4\right)-4 \cos \left(3 x^2+6 x-4\right)\right]+c\)
- 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 \(\frac{e^{4 x^2+8 x-4}}{50}\left[4 \sin \left(3 x^2+6 x-4\right)-3 \cos \left(3 x^2+6 x-4\right)\right]+c\)
Answer & Solution
Correct Answer
(B) \(\frac{e^{4 x^2+8 x-4}}{50}\left[4 \cos \left(3 x^2+6 x-4\right)+3 \sin \left(3 x^2+6 x-4\right)\right]+c\)
Step-by-step Solution
Detailed explanation
\(I=\int e^{4 x^2+8 x-4}(x+1) \cos \left(3 x^2+6 x-4\right) d x\) 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, \(\int e^{a x} \cos (b x+c) d x\) \(=\frac{e^{a x}}{a^2+b^2}(a \cos (b x+c)+b \sin (b x+c))\)…
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