Key to "Evaluate the following integrals. ∫t2sin⁡(3t)dt"

CheemnCatelvew

Answered question

2021-10-15

Evaluate the following integrals.

\int t^{2} \sin (3 t) d t

Fatema Sutton

Skilled2021-10-16Added 88 answers

Here,

\int t^{2} \sin (3 t) d t

= - \frac{t^{2}}{3} \cos (3 t) - \int (- \frac{\cos (3 t)}{3}) \times \frac{d}{d t} (t^{2}) d t

(By the method of by-parts)

= - \frac{t^{2}}{3} \cos (3 t) + \frac{2}{3} \int t \cos (3 t) d t

= - \frac{t^{2}}{3} \cos (3 t) + \frac{2}{3} [t \frac{\sin (3 t)}{3} - \int \frac{\sin (3 t)}{3} \times \frac{d}{d t} (t) d t]

= - \frac{t^{2}}{3} \cos (3 t) + \frac{2}{3} [t \frac{\sin (3 t)}{3} - \int \frac{\sin (3 t)}{3} d t]

= - \frac{t^{2}}{3} \cos (3 t) + \frac{2}{3} [t \frac{\sin (3 t)}{3} - (- \frac{\cos (3 t)}{9})] + c

(where c is arbitrary constant.)

= - \frac{t^{2}}{3} \cos (3 t) + \frac{2 t}{9} \sin (3 t) + \frac{2 \cos (3 t)}{27} + c

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