Correct Answer is : −x3cos⁡3x3+x2sin⁡3x3+2xcos⁡3x9-{x³ 3x}{3}+{x² 3x}{3}+2x 3x/9−3x3cos3x+3x2sin3x+92xcos3x −2sin⁡3x27+C-2 3x/27+C−272sin3x+C

-{x³ 3x}{3}-{x² 3x}{3}+2x 3x/9 -2 3x/27+C

{x³ 3x}{3}+{x² 3x}{3}+2x 3x/9 -2 3x/27+C

-{x³ 3x}{3}+{x² 3x}{3}+2x 3x/9 -2 3x/27+C

-{x³ 3x}{3}+{x² 3x}{3}-2x 3x/9 -2 3x/27+C

Correct Answer is : −x3cos⁡3x3+x2sin⁡3x3+2xcos⁡3x9-{x³ 3x}{3}+{x² 3x}{3}+2x 3x/9−3x3cos3x+3x2sin3x+92xcos3x −2sin⁡3x27+C-2 3x/27+C−272sin3x+C

Test Index

UPSC NDA Math Model Paper 2

Question : 98 of 120

Marks: +1, -0

\int x^{3} \sin 3 x dx =

$-\frac{x^{3} \cos 3x}{3}-\frac{x^{2} \sin 3x}{3}+\frac{2x \cos 3x}{9}$ $-\frac{2 \sin 3x}{27}+C$
$\frac{x^{3} \cos 3x}{3}+\frac{x^{2} \sin 3x}{3}+\frac{2x \cos 3x}{9}$ $-\frac{2 \sin 3x}{27}+C$
$-\frac{x^{3} \cos 3x}{3}+\frac{x^{2} \sin 3x}{3}+\frac{2x \cos 3x}{9}$ $-\frac{2 \sin 3x}{27}+C$
$-\frac{x^{3} \cos 3x}{3}+\frac{x^{2} \sin 3x}{3}-\frac{2x \cos 3x}{9}$ $-\frac{2 \sin 3x}{27}+C$

Solution:

Let,

I=\int x^{3} \sin 3 x dx

Using by part property

I=x^{3}\left[\frac{-\cos 3x}{3}\right]+\int 3x^{2}\cdot\frac{\cos 3x}{3}dx

I=\frac{-x^{3} \cos 3x}{3}+[x^{2} \int \cos 3x dx

-\int 2x \cdot \frac{\sin 3x}{3} dx]

I=\frac{-x^{3} \cos 3x}{3}+\frac{x^{2} \sin 3x}{3}

-\frac{2}{3} \int x \cdot \sin 3x dx

I=\frac{-x^{3} \cos 3x}{3}+\frac{x^{2} \sin 3x}{3}

-\frac{2}{3}\left[x \int \sin 3x dx+\int (1) \frac{\cos 3x}{3} dx\right]

I=-\frac{x^{3} \cos 3x}{3}+\frac{x^{2} \sin 3x}{3}+\frac{2x \cos 3x}{9}

-\frac{2 \sin 3x}{27}+C

Hence, option (3) is correct.

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