Let I=∫dx(5+4cos⁡x)2 Let A=sin⁡x(5+4cos⁡x) Differentiating wrt.x dAdx=(5+4cos⁡x)(cos⁡x)−sin⁡x(−4sin⁡x)(5+4cos⁡x)2dAdx=5cosx+45+4cosx2=544cosx+5+4−2545+4cosx2dAdx=541(5+4cos⁡x)−941(5+4cos⁡x)2 Now integrating both the sides wrt.x ∫dAdx=54∫15+4cos⁡xdx−94∫1(5+4cos⁡x)2dxA=54∫15+4cos⁡xdx−94I94I=54∫15+4cos⁡xdx−A equation −(1) Cosider ∫15+4cos⁡xdx Put tan⁡x2=tdx=2dt1+t2cos⁡x=1−t21+t2=∫15+41−t21+t22dt1+t2=∫2dt5+st2+4−4t2=2∫2dtt2+9=2∫dtt2+32=2⋅13tan−1⁡t3=23tan−1⁡tan⁡x23 equation −(2)Sub⁡(2)in(1)94I=54×23tan−1⁡tan⁡x23−sin⁡x(5+4cos⁡x)+C1I=56×49tan−1⁡tan⁡x23−49sin⁡x(5+4cos⁡x)+49C1I=1027tan−1⁡tan⁡x23−49sin⁡x5+4cos⁡x+C

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$\int \frac{d x}{{(5 + 4 \cos x)}^{2}} =$

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$\frac{1}{9} \tan^{- 1} (\frac{x}{3}) - (\frac{\sin x}{5 + 4 \cos x}) + c$

$\frac{1}{27} \tan^{- 1} (\tan \frac{x}{2}) - \frac{4}{9} (\frac{\sin x}{5 + 4 \cos x}) + c$

$\frac{10}{27} \tan^{- 1} (\frac{\tan \frac{x}{2}}{3}) - \frac{4}{9} (\frac{\sin x}{5 + 4 \cos x}) + c$

$\frac{20}{27} \tan^{- 1} (\frac{\tan \frac{x}{2}}{3}) - (\frac{\sin x}{5 + 4 \cos x}) + c$

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Detailed Solution

$\begin{array}{l} Let I = \int \frac{d x}{(5 + 4 \cos x)^{2}} \\ Let A = \frac{\sin x}{(5 + 4 \cos x)} \\ Differentiating wrt.x \\ \frac{d A}{d x} = \frac{(5 + 4 \cos x) (\cos x) - \sin x (- 4 \sin x)}{(5 + 4 \cos x)^{2}} \end{array}$

$\frac{d A}{d x} = \frac{5 \cos x + 4}{{(5 + 4 \cos x)}^{2}} = \frac{\frac{5}{4} (4 \cos x + 5) + 4 - \frac{25}{4}}{{(5 + 4 \cos x)}^{2}}$

$\begin{array}{l} \frac{d A}{d x} = \frac{5}{4} (\frac{1}{(5 + 4 \cos x)}) - \frac{9}{4} (\frac{1}{(5 + 4 \cos x)^{2}}) \\ Now integrating both the sides wrt.x \\ \int \frac{d A}{d x} = \frac{5}{4} \int \frac{1}{5 + 4 \cos x} d x - \frac{9}{4} \int \frac{1}{(5 + 4 \cos x)^{2}} d x \\ A = \frac{5}{4} \int \frac{1}{5 + 4 \cos x} d x - \frac{9}{4} I \\ \frac{9}{4} I = \frac{5}{4} \int \frac{1}{5 + 4 \cos x} d x - A equation - (1) \\ Cosider \int \frac{1}{5 + 4 \cos x} d x \end{array}$

$\begin{array}{l} Put \tan \frac{x}{2} = t \\ d x = \frac{2 d t}{1 + t^{2}} \\ \cos x = \frac{1 - t^{2}}{1 + t^{2}} \\ = \int \frac{1}{5 + 4 (\frac{1 - t^{2}}{1 + t^{2}})} \frac{2 d t}{1 + t^{2}} \\ = \int \frac{2 d t}{5 + s t^{2} + 4 - 4 t^{2}} \\ = 2 \int \frac{2 d t}{t^{2} + 9} \end{array}$

$\begin{array}{l} = 2 \int \frac{d t}{t^{2} + 3^{2}} \\ = 2 \cdot \frac{1}{3} \tan^{- 1} (\frac{t}{3}) \\ = \frac{2}{3} \tan^{- 1} (\frac{\tan \frac{x}{2}}{3}) equation - (2) \\ Sub (2) in (1) \end{array}$

$\begin{array}{l} \frac{9}{4} I = \frac{5}{4} \times \frac{2}{3} \tan^{- 1} (\frac{\tan \frac{x}{2}}{3}) - \frac{\sin x}{(5 + 4 \cos x)} + C_{1} \\ I = \frac{5}{6} \times \frac{4}{9} \tan^{- 1} (\frac{\tan \frac{x}{2}}{3}) - \frac{4}{9} \frac{\sin x}{(5 + 4 \cos x)} + \frac{4}{9} C_{1} \\ I = \frac{10}{27} \tan^{- 1} (\frac{\tan \frac{x}{2}}{3}) - \frac{4}{9} (\frac{\sin x}{5 + 4 \cos x}) + C \end{array}$