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# Solve the following integral:$\int {\dfrac{{dx}}{{\sin x + \sin 2x}}}$

Last updated date: 29th Mar 2023
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Hint- For solving such problems, try to convert the term in simpler form by the help of trigonometric identities ( sin 2x =2 sin x cos x ) and use substitution method and partial fraction.

To solve the given integral, first solving the denominator
$\sin x + \sin 2x = \sin x + 2\sin x\cos x = \sin x\left[ {1 + 2\cos x} \right]$
Now let’s put the fraction together and proceeding further:
$\dfrac{1}{{\sin x + \sin 2x}} = \dfrac{1}{{\sin x\left[ {1 + 2\cos x} \right]}} = \dfrac{{\sin x}}{{{{\sin }^2}x\left[ {1 + 2\cos x} \right]}}$
Now we can integrate the above term:
$\int {\dfrac{{\sin x}}{{{{\sin }^2}x\left[ {1 + 2\cos x} \right]}}dx}$
Let us consider
$u = \cos x \\ du = - \sin xdx \\ {\sin ^2}x = 1 - {\cos ^2}x = 1 - {u^2} \\ \int {\dfrac{1}{{\left( {1 - {u^2}} \right)\left( {1 + 2u} \right)}}} \left( { - du} \right) \\ \int {\dfrac{{ - 1}}{{\left( {1 - u} \right)\left( {1 + u} \right)\left( {1 + 2u} \right)}}} du \\$
Now let’s take a break from integration and decompose the expression into partial fractions:
$\dfrac{{ - 1}}{{\left( {1 - u} \right)\left( {1 + u} \right)\left( {1 + 2u} \right)}} = \dfrac{A}{{\left( {1 + u} \right)}} + \dfrac{B}{{\left( {1 - u} \right)}} + \dfrac{C}{{\left( {1 + 2u} \right)}}$
Multiplying on both sides $\left( {1 - u} \right)\left( {1 + u} \right)\left( {1 + 2u} \right)$to get rid of denominators. We get:
$- 1 = A\left( {1 - u} \right)\left( {1 + 2u} \right) + B\left( {1 + u} \right)\left( {1 + 2u} \right) + C\left( {1 - u} \right)\left( {1 + u} \right) \\ - 1 = A\left( {1 + u - 2{u^2}} \right) + B\left( {1 + 3u + 2{u^2}} \right) + C\left( {1 - {u^2}} \right) \\ - 1 = \left( { - 2A + 2B - C} \right){u^2} + \left( {A + 3B} \right)u + \left( {A + B + C} \right) \\$
And now on comparing coefficient from both sides, we have three equations and three variables, so solving further:
$0 = - 2A + 2B - C \\ 0 = A + 3B \\ -1 = A + B + C \\$
Adding all the three equations, we get:
$- 1 = 6B \\ B = \dfrac{{ - 1}}{6} \\$
From the second equation, we get:
$A = - 3B = \dfrac{3}{6} = \dfrac{1}{2}$
From the third equation, we get:
$C = - 1 - A - B = - 1 - \dfrac{1}{2} + \dfrac{1}{6} = \dfrac{{ - 4}}{3}$
And so we have:
$\dfrac{{ - 1}}{{\left( {1 - u} \right)\left( {1 + u} \right)\left( {1 + 2u} \right)}} = \dfrac{1}{{2\left( {1 + u} \right)}} - \dfrac{1}{{6\left( {1 - u} \right)}} - \dfrac{4}{{3\left( {1 + 2u} \right)}}$
Now we can continue with the integration:
${\text{ = }}\int {\dfrac{1}{{2\left( {1 + u} \right)}}du - \int {\dfrac{1}{{6\left( {1 - u} \right)}}du - \int {\dfrac{4}{{3\left( {1 + 2u} \right)}}du} } } \\ = \dfrac{1}{2}\ln \left| {1 + u} \right| + \dfrac{1}{6}\ln \left| {1 - u} \right| - \dfrac{2}{3}\ln \left| {1 + 2u} \right| + C{\text{ }}\left[ {\because \int {\dfrac{1}{x}dx = \ln x + C} } \right] \\$
Now after substituting back the considered value of $u$ we get:
${\text{ = }}\dfrac{1}{2}\ln \left| {1 + \cos x} \right| + \dfrac{1}{6}\ln \left| {1 - \cos x} \right| - \dfrac{2}{3}\ln \left| {1 + 2\cos x} \right| + C$

Note- Integration of some questions as above where direct integration is not possible, solving by the use of partial fraction is very useful. Also sometimes we have to consider some functions like $\sin \& \cos$in terms of some variables, such as $u$ in this case.