$\int e^{- 3 x} (x^{2} + \sin 4 x) d x =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\begin{aligned} - e^{- 3 x} (\frac{x^{2}}{3} + \frac{2 x}{9} + \frac{2}{27} + \frac{3}{25} \sin 4 x + \frac{4}{25} \cos 4 x) + c \end{aligned}$

$\begin{array}{r} - e^{- 3 x} (\frac{x^{2}}{3} - \frac{2 x}{9} + \frac{2}{27} + \frac{3}{25} \sin 4 x + \frac{4}{25} \cos 4 x) + c \end{array}$

$\begin{array}{r} - e^{- 3 x} (\frac{x^{2}}{3} + \frac{2 x}{9} + \frac{2}{27} + \frac{3}{25} \sin 4 x - \frac{4}{25} \cos 4 x) + c \end{array}$

$- e^{- 3 x} (\frac{x^{2}}{3} - \frac{2 x}{9} + \frac{2}{27} + \frac{3}{25} \sin 4 x - \frac{4}{25} \cos 4 x) + c$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \int e^{- 3 x} (x^{2} + \sin 4 x) d x = \int e^{- 3 x} x^{2} d x + \int e^{- 3 x} \sin 4 x (2 x \\ = x^{2} \frac{e^{- 3 x}}{- 3} - \int \frac{2 x e^{- 3 x}}{- 3} d x + \frac{e^{- 3 x}}{9 + 16} (- 3 \sin 4 x - 4 \cos 4 x) \\ = \frac{x^{2} e^{- 3 x}}{- 3} + \frac{2}{3} \int x e^{- 3 x} d x + \frac{e^{- 3 x}}{25} (- 3 \sin 4 x - 4 \cos 4 x) \\ = \frac{x^{2} e^{- 3 x}}{- 3} + \frac{2}{3} [x \frac{e^{- 3 x}}{- 3} - \int \frac{e^{- 3 x}}{- 3} d x] + \frac{e^{3 x}}{25} (- 3 \sin 4 x - 4 \cos 4 x) \\ = \frac{x^{2} e^{- 3 x}}{- 3} - \frac{2 x}{9} e^{- 3 x} + \frac{2}{9} \frac{e^{- 3 x}}{- 3} + \frac{e^{- 3 x}}{25} (- 3 \sin 4 x - 4 \cos 4 x) + c \\ = - e^{- 3 x} (\frac{x^{2}}{3} + \frac{2 x}{9} + \frac{2}{27} + \frac{3}{25} \sin 4 x + \frac{4}{25} \cos 4 x) + c \end{array}$