Question

integral of 3cos4x / 2sin2x

Answer 1

\[\int\frac{3cos4x}{2sin2x}dx\]\[=\frac{3}{2}\int\frac{(1-2sin^22x)}{sin2x}dx\]\[=\frac{3}{2}\int\frac{1}{sin2x}-\frac{2sin^22x}{sin2x}dx\]\[=\frac{3}{2}\int(\frac{1}{sin2x}-2sin2x)dx\]

Answer 2

note: \(cos4x =cos^2 2x - sin^2 2x\)

\( \frac{3}{2} \int \frac{ \cos^2 2x - \sin^2 2x}{\sin 2x} dx\)

=\( \frac{3}{2} [\int \frac{ \cos^2 2x}{\sin 2x} dx- \int \sin 2x dx]\)

=\( \frac{3}{2} [\int \cos^2 (2x)d(\cos 2x)- \int \sin 2x dx]\)

do you require further help?

Answer 3

@Shadowys The last step is wrong?

Answer 4

oh right. sorry.

Answer 5

d(cos2x) gives -sin2x instead of 1/sin2x