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Question

Mathematics Question on Integration

sin2x cosx dx=\int\sin2x\ cosx\ dx=

A

13cos3x+C\frac{-1}{3}cos^3x+C

B

23cos3x+C\frac{-2}{3}cos^3x+C

C

23cos3x+C\frac{2}{3}cos^3x+C

D

13cos3x+C\frac{1}{3}cos^3x+C

E

43cos3x+C\frac{-4}{3}cos^3x+C

Answer

23cos3x+C\frac{-2}{3}cos^3x+C

Explanation

Solution

The correct option is (B): 23cos3x+C\frac{-2}{3}cos^3x+C