2022TrialShort answerQ16 (b) (i)2 marksQuestionIf In=∫0π/2cosnθ dθ,I_n=\int_0^{\pi/2}\cos^n\theta\,d\theta,In=∫0π/2cosnθdθ, show that In=n−1nIn−2.I_n=\frac{n-1}{n}I_{n-2}.In=nn−1In−2.Try the question before revealing the answer.Show solution ↻Solution(b)(i) In=∫0π2cosnθ dθI_n=\displaystyle\int_0^{\frac{\pi}{2}}\cos^n\theta\,d\thetaIn=∫02πcosnθdθ u=cosn−1θu=\cos^{n-1}\thetau=cosn−1θ v=sinθv=\sin\thetav=sinθ u′=−(n−1)cosn−2θsinθu'=-(n-1)\cos^{n-2}\theta\sin\thetau′=−(n−1)cosn−2θsinθ v′=cosθv'=\cos\thetav′=cosθ By ∫uv′ dx=uv−∫vu′ dx\int uv'\,dx=uv-\int vu'\,dx∫uv′dx=uv−∫vu′dx ∫0π2cosnθ dθ=[cosn−1θsinθ]0π2+(n−1)∫0π2cosn−2θsin2θ dθ\int_0^{\frac{\pi}{2}}\cos^n\theta\,d\theta=\left[\cos^{n-1}\theta\sin\theta\right]_0^{\frac{\pi}{2}}+(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}\theta\sin^2\theta\,d\theta∫02πcosnθdθ=[cosn−1θsinθ]02π+(n−1)∫02πcosn−2θsin2θdθ In=0+(n−1)∫0π2cosn−2θ(1−cos2θ) dθI_n=0+(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}\theta(1-\cos^2\theta)\,d\thetaIn=0+(n−1)∫02πcosn−2θ(1−cos2θ)dθ In=(n−1)In−2−(n−1)InI_n=(n-1)I_{n-2}-(n-1)I_nIn=(n−1)In−2−(n−1)In nIn=(n−1)In−2⇒In=n−1nIn−2.nI_n=(n-1)I_{n-2}\quad\Rightarrow\quad I_n=\frac{n-1}{n}I_{n-2}.nIn=(n−1)In−2⇒In=nn−1In−2.Supplied 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