1/2 _e |1 + cos x| + 1/6 _e |1 - cos x| - 2/3 _e |1 + 2 cos x| + c

1/3 _e |1 + cos x| - 2/3 _e |1 - cos x| - 1/2 _e |1 + 2 cos x| + c

1/2 _e |1 + sin x| - 1/3 _e |1 - sin x| - 1/3 _e |1 + cos x| + c

1/3 _e |1 - sin x| + 1/2 _e |1 + cos x| - 2/3 _e |1 - 2 cos x| + c

Test Index

AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 74 of 160

Marks: +1, -0

∫

\frac{dx}{\sin x+\sin 2x}

$\frac{1}{2} \log_e$ |1 + cos x| + $\frac{1}{6} \log_e$ |1 - cos x| - $\frac{2}{3} \log_e$ |1 + 2 cos x| + c
$\frac{1}{3} \log_e$ |1 + cos x| - $\frac{2}{3} \log_e$ |1 - cos x| - $\frac{1}{2} \log_e$ |1 + 2 cos x| + c
$\frac{1}{2} \log_e$ |1 + sin x| - $\frac{1}{3} \log_e$ |1 - sin x| - $\frac{1}{3} \log_e$ |1 + cos x| + c
$\frac{1}{3} \log_e$ |1 - sin x| + $\frac{1}{2} \log_e$ |1 + cos x| - $\frac{2}{3} \log_e$ |1 - 2 cos x| + c

Solution:

Consider the given expression,

\int\;\frac{dx}{\sin x+\sin 2x} \;\; =\int\;\frac{dx}{\sin x+\sin 2x}

\;\; =\int\;\frac{dx}{\sin x(1+2 \cos x)}

\;\; =\int\;\frac{\sin x \, dx}{\sin ^{2} x(1+2 \cos x)}

\;\; =\int\;\frac{-\sin x \, dx}{(\cos ^{2} x-1)(1+2 \cos x)}

Let

\cos x=t

then

-\sin x \, dx = dt

\text{So}

\int\;\frac{dt}{(t^{2}-1)(1+2 t)}

By partial fraction

A = \frac{1}{6}, B = \frac{1}{2} \;\text{ and }\; C = -\frac{4}{3}

Now,

\int\;\frac{dt}{(t^{2}-1)(1+2 t)} \;\; =\int\;\frac{\frac{1}{6}}{t-1} dt + \frac{1}{2} \int\;\frac{dt}{t+1} - \frac{4}{6} \int\;\frac{dt}{t+\frac{1}{2}}

\;\; =\frac{\ln(t-1)}{6}+\frac{1}{2} \ln(t+1)-\frac{4}{6} \ln\left(t+\frac{1}{2}\right)+C

\;\; =\left[ \begin{array}{l} \frac{\ln(\cos x-1)}{6}+\frac{1}{2} \ln(\cos x+1) \\ -\frac{2}{3} \ln\left(\cos x+\frac{1}{2}\right)+C \end{array} \right].

\;\; =\left[ \begin{array}{l} \frac{1}{2} \ln|1+\cos x|+\frac{1}{6} \ln|1-\cos x| \\ -\frac{2}{3} \ln\left|\cos x+\frac{1}{2}\right|+C \end{array} \right].

Further simplify the above expression

\int\;\frac{dt}{(t^{2}-1)(1+2 t)}

= \frac{1}{2} \ln(1+\cos x)+\frac{1}{6} \ln|1-\cos x|-\frac{2}{3} \ln|1+2 \cos x|+C

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