Question

Find the integral \(\int sin⁡2x+e^3x-cos⁡3x dx\).

(a) –\(\frac{sin⁡2x}{2}+\frac{e^{3x}}{3}-\frac{sin⁡3x}{3}+C\)

(b) –\(\frac{cos⁡2x}{2}+\frac{e^{3x}}{3}-\frac{sin⁡3x}{3}+C\)

(c) \(\frac{cos⁡2x}{2}+\frac{e^{3x}}{3}-\frac{cos⁡3x}{3}+C\)

(d) –\(\frac{cos⁡2x}{2}-\frac{e^{3x}}{3}+\frac{cos⁡3x}{3}+C\)

This question was addressed to me in an internship interview.

I would like to ask this question from Integration as an Inverse Process of Differentiation in portion Integrals of Mathematics – Class 12

Answer 1

The correct option is (b) –\(\frac{cos⁡2x}{2}+\frac{e^{3x}}{3}-\frac{sin⁡3x}{3}+C\)

To explain I would say: To find \(\int \,sin⁡2x+e^{3x}-cos⁡3x \,dx\)

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

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