Question

Integrate 5x sin⁡3x.

(a) –\(\frac{5}{3} \,x \,cos⁡3x+\frac{5}{9} tan⁡3x+C\)

(b) \(\frac{5}{3} \,cos⁡3x-\frac{5}{9} \,sin⁡3x+C\)

(c) \(x cos⁡3x+\frac{5}{9} \,sin⁡3x+C\)

(d) –\(\frac{5}{3} \,x \,cos⁡3x+\frac{5}{9} \,sin⁡3x+C\)

I got this question in an internship interview.

The question is from Integration by Parts in portion Integrals of Mathematics – Class 12

Answer 1

Right answer is (d) –\(\frac{5}{3} \,x \,cos⁡3x+\frac{5}{9} \,sin⁡3x+C\)

Easy explanation: By using the formula, ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) ,we get

∫ 5x sin⁡3x dx=5x∫ sin⁡3x dx -∫ (5x)’∫ sin⁡3x dx

=\(5x(-\frac{cos⁡3x}{3})+\int \frac{5 cos⁡3x}{3} dx\)

=-\(\frac{5}{3} x cos⁡3x+\frac{5}{9} sin⁡3x+C\)