Question

Solve the following

$\int \: \frac{dx}{sinx + sin2x}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{sinx + sin2x}$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{sinx + 2 \: sinx \: cosx}$

$\rm \:  =  \: \displaystyle\int\rm \frac{dx}{sinx(1 + 2 \: cosx)}$

$\rm \:  =  \: \displaystyle\int\rm \frac{sinx}{sin^{2} x(1 + 2 \: cosx)} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{sinx}{(1 - cos^{2} x)(1 + 2 \: cosx)} \: dx$

Now, to evaluate this integral further, we use Method of Substitution

So, Substitute

$\purple{\rm :\longmapsto\:cosx = y}$

$\purple{\rm :\longmapsto\: - sinx \: dx \: = \: dy}$

$\purple{\rm :\longmapsto\:sinx \: dx \: = \: - \: dy}$

So, on substituting these values, we get

$\rm \:  =  \: \displaystyle\int\rm \frac{ - dy}{(1 - {y}^{2})(1 + 2y)}$

$\rm \:  =  \: \displaystyle\int\rm \frac{dy}{({y}^{2} - 1)(1 + 2y)}$

$\rm \:  =  \: \displaystyle\int\rm \frac{dy}{({y}- 1)(y + 1)(1 + 2y)}$

Now, to solve it further, we use method of Partial Fractions.

So, Let assume that

$\rm :\longmapsto\:\dfrac{1}{(y + 1)(y - 1)(2y + 1)} = \dfrac{a}{y + 1} + \dfrac{b}{y - 1} + \dfrac{c}{2y + 1} - - - (1)$

On taking LCM, we get

$\rm :\longmapsto\:1 = a(y - 1)(2y + 1) + b(y + 1)(2y + 1) + c(y - 1)(y + 1)$

On substituting y = - 1, we get

$\rm :\longmapsto\:1 = a( - 1 - 1)( - 2 + 1) +0 + 0$

$\rm :\longmapsto\:1 = a( - 2)( - 1)$

$\bf\implies \:a = \dfrac{1}{2}$

On substituting y = 1, we get

$\rm :\longmapsto\:1 = 0 + b(1+ 1)(2+ 1) + 0$

$\rm :\longmapsto\:1 = b(2)(3)$

$\bf\implies \:b = \dfrac{1}{6}$

On substituting y = - 1/2, we get

$\rm :\longmapsto\:1 = 0 + 0+ c\bigg( - \dfrac{1}{2} - 1\bigg)\bigg( - \dfrac{1}{2} + 1\bigg)$

$\rm :\longmapsto\:1 = c\bigg( - \dfrac{3}{2}\bigg)\bigg(\dfrac{1}{2}\bigg)$

$\bf\implies \:c = \: - \dfrac{4}{3}$

So, equation (1) can be rewritten as

$\rm :\longmapsto\:\dfrac{1}{(y + 1)(y - 1)(2y + 1)} = \dfrac{1}{2(y + 1)} + \dfrac{1}{6(y - 1)} - \dfrac{4}{3(2y + 1)}$

So, on integrating both sides w. r. t. y, we get

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{dy}{(y + 1)(y - 1)(2y + 1)} =\displaystyle\int\rm \dfrac{dy}{2(y + 1)} + \displaystyle\int\rm \dfrac{dy}{6(y - 1)} - \displaystyle\int\rm \dfrac{4dy}{3(2y + 1)}$

$\rm \:  =  \: \dfrac{1}{2} log |y + 1| + \dfrac{1}{6}log |y - 1| - \dfrac{4}{3} \times \dfrac{log |2y + 1| }{2} + c$

$\rm \:  =  \: \dfrac{1}{2} log |y + 1| + \dfrac{1}{6}log |y - 1| - \dfrac{2}{3}log |2y + 1| + c$

$\rm \:  =  \: \dfrac{1}{2} log |cosx + 1| + \dfrac{1}{6}log |cosx - 1| - \dfrac{2}{3}log |2cosx + 1| + c$

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LEARN MORE

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Fraction & \bf Partial \: Fraction \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \dfrac{1}{(ax + b)(cx + d)} & \sf \dfrac{p}{ax + b} + \dfrac{q}{cx + d} \\ \\ \sf \dfrac{1}{ {(ax + b)}^{2} } & \sf \dfrac{p}{ax + b} + \dfrac{q}{ {(ax + b)}^{2} } \\ \\ \sf \dfrac{1}{(ax + b)(c {x}^{2} + d)} & \sf \dfrac{p}{ax + b} + \dfrac{qx + r}{c {x}^{2} + d} \end{array}} \\ \end{gathered}$

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$