Trig. integration

[stuart clark]

[tex]\bf{\displaystyle \int\frac{1}{(3+4\cos x)^2}dx}[/tex]

[stuart clark]

Thanks friends got it

Let [tex]\displaystyle\bf{I=\frac{\sin x}{5+4\cos x}}\;[/tex], Then


[tex]\displaystyle\bf{\frac{dI}{dx}=\frac{(5+4\cos x).(\cos x)-\sin x.(-4\sin x)}{(5+4\cos x)^2}}[/tex]


[tex]\displaystyle\bf{\frac{dI}{dx}=\frac{5\cos x+4}{(5+4\cos x)^2}=\frac{5}{4}.\frac{(4\cos x+5)}{(5+4\cos x)^2}+\left(4-\frac{25}{4}\right).\frac{1}{(5+4\cos x)^2}}[/tex]


[tex]\displaystyle\bf{\frac{dI}{dx}=\frac{5}{4}.\frac{1}{(5+4\cos x)}-\frac{9}{4}.\frac{1}{(5+4\cos x)^2}}[/tex]


[tex]\displaystyle\bf{\int\frac{dI}{dx}dx=\frac{5}{4}\int\frac{1}{(5+4\cos x)}dx-\frac{9}{4}\int\frac{1}{(5+4\cos x)^2}dx}[/tex]


[tex]\displaystyle\bf{\int\frac{1}{(5+4\cos x)^2}dx = \frac{5}{9}\int\frac{1}{5+4\cos x}dx-\frac{4}{9}.I}[/tex]


Now Let [tex]\displaystyle\bf{J=\int\frac{1}{5+4\cos x}dx}[/tex]


Put [tex]\displaystyle\bf{\cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}}[/tex]


[tex]\displaystyle\bf{J=\int\frac{\sec^2 \frac{x}{2}}{9+\tan^2 \frac{x}{2}}dx}[/tex]


Now Put [tex]\displaystyle\bf{\tan \frac{x}{2}=t\Leftrightarrow \sec^2 \frac{x}{2}dx = 2tdt}[/tex]


[tex]\displaystyle\bf{J=2\int\frac{1}{3^2+t^2}dt = \frac{2}{3}\tan^{-1}\left(\frac{t}{3}\right)}[/tex]


[tex]\displaystyle\bf{J=\frac{2}{3}\tan^{-1}\left(\frac{\tan \frac{x}{2}}{3}\right)}[/tex]


So [tex]\displaystyle\bf{\int\frac{1}{(5+4\cos x)}dx = \frac{10}{27}\tan^{-1}\left(\frac{\tan \frac{x}{2}}{3}\right)-\frac{4}{9}\left(\frac{\sin x}{5+4\cos x}\right)+C}[/tex]