2022TrialShort answerQ12 (d) (i)2 marksQuestionUsing the substitution x=secθx=\sec\thetax=secθ or otherwise show by integrating that ∫1x2−1 dx=ln(x+x2−1).\int\frac{1}{\sqrt{x^2-1}}\,dx=\ln\left(x+\sqrt{x^2-1}\right).∫x2−11dx=ln(x+x2−1).Try the question before revealing the answer.Show solution ↻Solution∫1x2−1 dx=ln(x+x2−1)+C\int\frac{1}{\sqrt{x^2-1}}\,dx=\ln\left(x+\sqrt{x^2-1}\right)+C∫x2−11dx=ln(x+x2−1)+C Working Out: x=secθx=\sec\thetax=secθ dx=secθtanθ dθdx=\sec\theta\tan\theta\,d\thetadx=secθtanθdθ =∫1tanθ⋅secθtanθ dθ=\int\frac{1}{\tan\theta}\cdot\sec\theta\tan\theta\,d\theta=∫tanθ1⋅secθtanθdθ =∫secθ dθ=\int\sec\theta\,d\theta=∫secθdθ =∫sec2θ+secθtanθsecθ+tanθ dθ=\int\frac{\sec^2\theta+\sec\theta\tan\theta}{\sec\theta+\tan\theta}\,d\theta=∫secθ+tanθsec2θ+secθtanθdθ =ln[secθ+tanθ]+C=\ln\left[\sec\theta+\tan\theta\right]+C=ln[secθ+tanθ]+C =ln[x+x2−1]+C=\ln\left[x+\sqrt{x^2-1}\right]+C=ln[x+x2−1]+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