Đáp án:
${{2\pi } \over {15}} - {1 \over 5}\ln {{1 + 2\sqrt 3 } \over 2}$
Giải thích các bước giải:
Ta có:
$\eqalign{
& \int\limits_0^{{\pi \over 3}} {{{\sin x} \over {\cos x + 2\sin x}}} dx \cr
& = \int\limits_0^{{\pi \over 3}} {{{{2 \over 5}(\cos x + 2\sin x)} \over {\cos x + 2\sin x}}} dx - \int\limits_0^{{\pi \over 3}} {{{{1 \over 5}(2\cos x - \sin x)} \over {\cos x + 2\sin x}}dx} \cr
& = {2 \over 5}\int\limits_0^{{\pi \over 3}} {dx} - {1 \over 5}\int\limits_0^{{\pi \over 3}} {{{d(2\sin x + \cos x)} \over {\cos x + 2\sin x}}} \cr
& = {2 \over 5}x\left| {\matrix{
{{\pi \over 3}} \cr
0 \cr
} - {1 \over 5}\ln \left| {\cos x + 2\sin x} \right|\left| {\matrix{
{{\pi \over 3}} \cr
0 \cr
} } \right.} \right. \cr
& = {2 \over 5}.{\pi \over 3} - {1 \over 5}\ln \left| {{1 \over 2} + \sqrt 3 } \right| + {1 \over 5}\ln \left| {1 + 0} \right| \cr
& = {{2\pi } \over {15}} - {1 \over 5}\ln {{1 + 2\sqrt 3 } \over 2} \cr} $
