G
Guest
Guest
Hi
I'm trying to get to grips with intergrating trig functions and I would be grateful if you would look over some of my work to make sure I'm doing it right. Thank You.
\(\displaystyle \L
\int {\frac{{2\tan x}}{{\sin 2x}}dx}
= \int {\frac{{2\sin x}}{{2\cos x\cos x\sin x}}} dx
= \int {\frac{1}{{\cos ^2 x}}} dx\)
\(\displaystyle \left[ {\cos ^2 x = {\textstyle{1 \over 2}}(1 + \cos 2x) = {\textstyle{1 \over 2}} + {\textstyle{1 \over 2}}\cos 2x} \right]\)
\(\displaystyle \L
\begin{array}{l}
\Rightarrow \int {(2 + 2\cos 2x)dx = 2x + \sin 2x + c} \\
\\
\end{array}\)
:?:
I'm trying to get to grips with intergrating trig functions and I would be grateful if you would look over some of my work to make sure I'm doing it right. Thank You.
\(\displaystyle \L
\int {\frac{{2\tan x}}{{\sin 2x}}dx}
= \int {\frac{{2\sin x}}{{2\cos x\cos x\sin x}}} dx
= \int {\frac{1}{{\cos ^2 x}}} dx\)
\(\displaystyle \left[ {\cos ^2 x = {\textstyle{1 \over 2}}(1 + \cos 2x) = {\textstyle{1 \over 2}} + {\textstyle{1 \over 2}}\cos 2x} \right]\)
\(\displaystyle \L
\begin{array}{l}
\Rightarrow \int {(2 + 2\cos 2x)dx = 2x + \sin 2x + c} \\
\\
\end{array}\)
:?:
