Q113 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
If $\int x^3 \sin 3 x d x=f(x) \cos 3 x+g(x) \sin 3 x+c$ then, $27(f(x)+x g(x))=$
- A$18 x^3+4 x$
- B$8 x$
- Correct answerC$4 x$
- D$18 x^3+8 x$
Solution
$\int x^3 \sin 3 x d x=x^3\left(\frac{-\cos 3 x}{3}\right)+\int x^2 \cos 3 x d x$
$\begin{aligned} & =-x^3 \frac{\cos 3 x}{3}+x^2 \frac{\sin 3 x}{3}-\frac{2}{3} \int(x \sin 3 x) d x \\ & =-x^3 \frac{\cos 3 x}{3}+\frac{x^2 \sin 3 x}{3}-\frac{2}{3}\left[\frac{-x \cos 3 x}{3}+\int \frac{\cos 3 x}{3} d x\right] \\ & =-x^3 \frac{\cos 3 x}{3}+x^2 \frac{\sin 3 x}{3}+2 x \frac{\cos 3 x}{9}-\frac{2}{27} \sin 3 x \\ & =\cos 3 x\left\{\frac{-x^3}{3}+\frac{2 x}{9}\right\}+\sin 3 x\left(\frac{x^2}{3}-\frac{2}{27}\right) \\ & \Rightarrow f(x)=\frac{-x^3}{3}+\frac{2 x}{9} ; g(x)=\frac{x^2}{3}-\frac{2}{27} \\ & \Rightarrow 27 \cdot\left(\frac{-x^3}{3}+\frac{2 x}{9}+\frac{x^3}{3}-\frac{2 x}{27}\right)=27 \times \frac{4 x}{27}=4 x\end{aligned}$
Hence, the correct answer is option: 'C'.