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Q.Obtain the reduction formula for In=∫Cosnx dxI_n = \int Cos^n x\, dx, nn being a positive integer n≥2n\geq 2, and deduce the value of ∫Cos3x dx\int Cos^3 x\, dx.

Telangana TsbieTelangana Board of Intermediate Education 2019Subjective· 7mImportance★★★★★
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Split cos⁡nx=cos⁡n−1x⋅cos⁡x\cos^n x=\cos^{n-1}x\cdot\cos x and integrate by parts, using sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x to relate InI_n back to itself and to In−2I_{n-2}.

Let In=∫cos⁡nx dxI_n=\displaystyle\int\cos^n x\,dx, n≥2n\geq2.

Write cos⁡nx=cos⁡n−1x⋅cos⁡x\cos^n x=\cos^{n-1}x\cdot\cos x and integrate by parts with u=cos⁡n−1xu=\cos^{n-1}x, dv=cos⁡x dxdv=\cos x\,dx, so du=−(n−1)cos⁡n−2xsin⁡x dxdu=-(n-1)\cos^{n-2}x\sin x\,dx, v=sin⁡xv=\sin x:

In=cos⁡n−1xsin⁡x+(n−1)∫cos⁡n−2xsin⁡2x dxI_n=\cos^{n-1}x\sin x+(n-1)\displaystyle\int\cos^{n-2}x\sin^2x\,dx

=cos⁡n−1xsin⁡x+(n−1)∫cos⁡n−2x(1−cos⁡2x) dx=\cos^{n-1}x\sin x+(n-1)\displaystyle\int\cos^{n-2}x(1-\cos^2x)\,dx

=cos⁡n−1xsin⁡x+(n−1)In−2−(n−1)In=\cos^{n-1}x\sin x+(n-1)I_{n-2}-(n-1)I_n

Bring the InI_n terms together:

In+(n−1)In=cos⁡n−1xsin⁡x+(n−1)In−2I_n+(n-1)I_n=\cos^{n-1}x\sin x+(n-1)I_{n-2}

nIn=cos⁡n−1xsin⁡x+(n−1)In−2nI_n=\cos^{n-1}x\sin x+(n-1)I_{n-2}

In=cos⁡n−1xsin⁡xn+n−1nIn−2I_n=\dfrac{\cos^{n-1}x\sin x}{n}+\dfrac{n-1}{n}I_{n-2}

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