TS EAMCET · Maths · Indefinite Integration
\(\int e^{-3 x}\left(x^2+\sin 4 x\right) d x=\)
- A \(-e^{-3 x}\left(\frac{x^2}{3}+\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x+\frac{4}{25} \cos 4 x\right)+C\)
- B \(-e^{-3 x}\left(\frac{x^2}{3}-\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x+\frac{4}{25} \cos 4 x\right)+C\)
- C \(-e^{-3 x}\left(\frac{x^2}{3}+\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x-\frac{4}{25} \cos 4 x\right)+C\)
- D \(-e^{-3 x}\left(\frac{x^2}{3}-\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x-\frac{4}{25} \cos 4 x\right)+C\)
Answer & Solution
Correct Answer
(A) \(-e^{-3 x}\left(\frac{x^2}{3}+\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x+\frac{4}{25} \cos 4 x\right)+C\)
Step-by-step Solution
Detailed explanation
\begin{aligned} \text { Let } I & =\int e^{-3 x}\left(x^2+\sin 4 x\right) d x \\ & \begin{aligned} & \int e^{-3 x} x^2 d x+\int e^{-3 x} \sin 4 x d x \\ &=x^2 \cdot \frac{e^{-3 x}}{-3}-\int-\frac{1}{3} e^{-3 x} \cdot 2 x d x+\frac{e^{-3 x}}{9+16} \\ &=-\frac{1}{3} x^2 e^{-3…
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