Question

Find ∫ 2 sin^3⁡x+1 dx

(a) \(\frac{3}{2}-\frac{cos⁡3x}{6}+x+C\)

(b) –\(\frac{3}{2} cos⁡x+\frac{cos⁡3x}{6}+x+C\)

(c) –\(\frac{3}{2} cos⁡x-\frac{cos⁡3x}{6}-x+C\)

(d) –\(\frac{3}{2} cos⁡x+\frac{cos⁡3x}{6}+C\)

I got this question during an interview for a job.

My doubt is from Methods of Integration-2 in section Integrals of Mathematics – Class 12

Answer 1

The correct answer is (b) –\(\frac{3}{2} cos⁡x+\frac{cos⁡3x}{6}+x+C\)

The best explanation: We know that, sin⁡3x=3 sin⁡x-4 sin^3⁡x

∴sin^3x=\(\frac{3 sin⁡x-sin⁡3x}{4}\)

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

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

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