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Question 68 of 96

Q.Evaluate: ∫cos⁡5x dx\displaystyle\int \cos^5 x \, dx

Puducherry TnboardTamil Nadu HSC (DGE) Board 2018Subjective· 6mImportance★★★★★
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Peel off one cosine factor, convert the remaining even power via cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x, and integrate term-by-term after the substitution s=sin⁡xs=\sin x.

  1. Write cos⁡5x=cos⁡4x⋅cos⁡x=(cos⁡2x)2cos⁡x=(1−sin⁡2x)2cos⁡x\cos^5x=\cos^4x\cdot\cos x=(\cos^2x)^2\cos x=(1-\sin^2x)^2\cos x.
  2. So ∫cos⁡5x dx=∫(1−sin⁡2x)2cos⁡x dx\displaystyle\int\cos^5x\,dx=\int(1-\sin^2x)^2\cos x\,dx.
  3. Substitute s=sin⁡xs=\sin x, so ds=cos⁡x dxds=\cos x\,dx: the integral becomes ∫(1−s2)2 ds\displaystyle\int(1-s^2)^2\,ds.
  4. Expand: (1−s2)2=1−2s2+s4(1-s^2)^2=1-2s^2+s^4. …

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