2022TrialShort answerQ12 (b)3 marksQuestionFind ∫12+cosx dx\displaystyle\int\frac{1}{2+\cos x}\,dx∫2+cosx1dx.Try the question before revealing the answer.Show solution ↻Solution∫12+cosx dx=23tan−1(tan(x/2)3)+C\int\frac{1}{2+\cos x}\,dx=\frac{2}{\sqrt3}\tan^{-1}\left(\frac{\tan(x/2)}{\sqrt3}\right)+C∫2+cosx1dx=32tan−1(3tan(x/2))+C Working Out: ∫12+cosx dx\quad \int\frac{1}{2+\cos x}\,dx∫2+cosx1dx t=tan(x2)t=\tan\left(\frac{x}{2}\right)t=tan(2x) dtdx=12sec2(x2)=1+t22\frac{dt}{dx}=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)=\frac{1+t^2}{2}dxdt=21sec2(2x)=21+t2 ∴dx=21+t2 dt\therefore dx=\frac{2}{1+t^2}\,dt∴dx=1+t22dt ∫12+1−t21+t2⋅21+t2 dt\int\frac{1}{2+\frac{1-t^2}{1+t^2}}\cdot\frac{2}{1+t^2}\,dt∫2+1+t21−t21⋅1+t22dt =∫23+t2 dt=\int\frac{2}{3+t^2}\,dt=∫3+t22dt =23tan−1(t3)+C=\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{t}{\sqrt{3}}\right)+C=32tan−1(3t)+C =23tan−1[tan(x2)3]+C=\frac{2}{\sqrt{3}}\tan^{-1}\left[\frac{\tan\left(\frac{x}{2}\right)}{\sqrt{3}}\right]+C=32tan−1[3tan(2x)]+CSupplied marking guidance · human reviewedCompare this with your working, not only your final sentence.↻ View questionSyllabus alignmentCalculusFurther IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.Spot a problem? Send feedbackWebsiteWhat needs attention?Question textSolutionDiagram or visualSomething elseTell us what looks wrongPlease do not include your name, school, or other personal details.Send for review