Evaluate the integral. \int \sin 3x\cos 2x dx.

jernplate8

jernplate8

Answered question

2021-11-06

Evaluate the integral.
sin3xcos2xdx.

Answer & Explanation

ottcomn

ottcomn

Skilled2021-11-07Added 97 answers

Step 1
The given integral is sin3xcos2xdx.
Step 2
sin3xcos2xdx=sin(3x+2x)+sin(3x2x)2dx
=12sin(5x)+sin(x)dx
=12[15cos(5x)cosx]+C
=12[15cos(5x)+cosx]+C
Hence, the value of the integral is 12[15cos(5x)+cosx]+C.

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