2022TrialShort answerQ12 (a)3 marksQuestionFind ∫sin−1(x) dx\displaystyle\int\sin^{-1}(x)\,dx∫sin−1(x)dx.Try the question before revealing the answer.Show solution ↻Solution∫sin−1(x) dx=xsin−1(x)+1−x2+C\int\sin^{-1}(x)\,dx=x\sin^{-1}(x)+\sqrt{1-x^2}+C∫sin−1(x)dx=xsin−1(x)+1−x2+C Working Out: u=sin−1(x),v′=1u=\sin^{-1}(x),\quad v'=1u=sin−1(x),v′=1 u′=11−x2,v=xu'=\frac{1}{\sqrt{1-x^2}},\quad v=xu′=1−x21,v=x ∫uv′ dx=uv−∫vu′ dx\int uv'\,dx=uv-\int vu'\,dx∫uv′dx=uv−∫vu′dx ∫sin−1(x) dx=xsin−1(x)−∫x1−x2 dx\int\sin^{-1}(x)\,dx=x\sin^{-1}(x)-\int\frac{x}{\sqrt{1-x^2}}\,dx∫sin−1(x)dx=xsin−1(x)−∫1−x2xdx =xsin−1(x)+12∫−2x1−x2 dx=x\sin^{-1}(x)+\frac{1}{2}\int\frac{-2x}{\sqrt{1-x^2}}\,dx=xsin−1(x)+21∫1−x2−2xdx =xsin−1(x)+12×21−x2+C=x\sin^{-1}(x)+\frac{1}{2}\times2\sqrt{1-x^2}+C=xsin−1(x)+21×21−x2+C =xsin−1(x)+1−x2+C=x\sin^{-1}(x)+\sqrt{1-x^2}+C=xsin−1(x)+1−x2+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