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Q.Obtain the reduction formula for ∫sin⁡nx dx\int \sin^n x\, dx for an integer n≥2n \ge 2 and deduce the value of ∫sin⁡4x dx\int \sin^4 x\, dx.

Telangana TsbieTelangana Board of Intermediate Education 2020Subjective· 7mImportance★★★★★
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Split sin⁡nx=sin⁡n−1x⋅sin⁡x\sin^n x=\sin^{n-1}x\cdot\sin x, integrate by parts, and use cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x to relate InI_n to In−2I_{n-2}.

Let In=∫sin⁡nx dxI_n=\displaystyle\int \sin^n x\,dx (n≥2n\ge2 an integer).

Write sin⁡nx=sin⁡n−1x⋅sin⁡x\sin^n x = \sin^{n-1}x\cdot\sin x. Integrate by parts with u=sin⁡n−1x, dv=sin⁡x dxu=\sin^{n-1}x,\ dv=\sin x\,dx, so du=(n−1)sin⁡n−2xcos⁡x dx, v=−cos⁡xdu=(n-1)\sin^{n-2}x\cos x\,dx,\ v=-\cos x:

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

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

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

Collect InI_n terms:

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

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

Reduction formula:

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

Deducing ∫sin⁡4x dx\int\sin^4x\,dx: apply with n=4n=4, then n=2n=2.

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