Question

\int \frac{2-3sin2x}{cos2x}dx

Answer 1

any idea?

Answer 2

\[let I=\int \frac{ 2-sin 3x }{ cos 2x }dx\]
\[=\int\limits \frac{ 2 }{ \cos ~2x }dx-3 \int\limits \frac{ \sin ~2x }{ \cos~2x }dx+c=I_{1}+I_{2}+c\]
\[I_{1}=\int\limits \frac{ 2 }{ \cos ~2x}dx\]
\[=\int\limits \frac{ 2 }{ \sin (\frac{ \pi }{ 2 }+2x) }dx=\int\limits \frac{ 2 }{ 2 \sin (\frac{ \pi }{ 4 }+x)\cos (\frac{ \pi }{ 4 } +x) }dx\]
divide the numerator and denominator by cos^2(\pi/4+x)
\[=\int\limits \frac{ \frac{ 1 }{ \cos^2(\frac{ \pi }{ 4 }+x) } }{ \tan (\frac{ \pi }{ 4 }+x) }dx\]
\[= \int\limits \frac{ \sec^2(\frac{ \pi }{ 4 } +x)}{ \tan (\frac{ \pi }{ 4 }+x) }dx\]
\[=\log \left| \tan(\frac{ \pi }{ 4 }+x) \right|\]
\[I_{2}=3 \int\limits \frac{ \sin~2x }{ \cos~2x }dx\]
put cos 2x=t
diff.
-2sin 2x dx=dt
\[\sin 2x dx=-\frac{ dt }{ 2 }\]
\[I_{2}=-\frac{ 3 }{ 2 } \int\limits \frac{ dt }{ t }=-\frac{ 3 }{ 2 }\log \left| t \right|=-\frac{ 3 }{2 }\log \left| \cos 2x \right|\]
complete it.