Question

Integrate sin^3⁡(x+2).

(a) \(\frac{3}{4} \,(sin⁡(x+2))+\frac{1}{12} \,cos⁡(3x+6)+C\)

(b) –\(\frac{3}{4} \,(cos⁡(x+2))-\frac{1}{5} \,cos⁡(3x+6)+C\)

(c) –\(\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,cos⁡(3x+6)+C\)

(d) –\(\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,sin⁡(x+2)+C\)

The question was posed to me during an interview.

Enquiry is from Methods of Integration-2 topic in chapter Integrals of Mathematics – Class 12

Answer 1

Right option is (c) –\(\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,cos⁡(3x+6)+C\)

Explanation: To find: ∫ 3 sin^3⁡(x+2) dx

We know that, sin⁡3x=3 sin⁡x-4 sin^3⁡x

∴sin^3⁡⁡x=\(\frac{3 sin⁡x-sin⁡3x}{4}\)

sin^3⁡(x+2)=\(\frac{(3 sin⁡(x+2)-sin⁡(3x+6))}{4}\)

\(\int sin^3⁡(x+2) \,dx=\frac{3}{4} \int sin⁡(x+2) \,dx-\frac{1}{4} \int \,sin⁡(3x+6) \,dx\)

=-\(\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,cos⁡(3x+6)+C\)