Question

Integrate 2x sin⁡2x.

(a) \(\frac{sin⁡2x}{2}\)+x cos⁡2x+C

(b) \(\frac{sin⁡2x}{2}\)-cos⁡2x+C

(c) \(\frac{cos⁡2x}{2}\)-x cos⁡2x+C

(d) \(\frac{sin⁡2x}{2}\)-x cos⁡2x+C

The question was asked during an interview.

The above asked question is from Integration by Parts in chapter Integrals of Mathematics – Class 12

Answer 1

Right choice is (d) \(\frac{sin⁡2x}{2}\)-x cos⁡2x+C

To explain: By using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx)

∫ x tanx dx=2x ∫ sin⁡2x dx-∫ (2x)’ ∫ sin⁡2x dx

=2x (-\(\frac{cos⁡2x}{2}\))+2\(\int \frac{cos⁡2x}{2}\) dx

=-xcos 2x+∫ cos⁡2x dx

=\(\frac{sin⁡2x}{2}\)-x cos⁡2x+C