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

Andhra Pradesh BieapBIEAP Intermediate Board 2024Subjective· 7mImportance★★★★★
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Integrate sin⁡nx=sin⁡n−1x⋅sin⁡x\sin^n x=\sin^{n-1}x\cdot\sin x by parts to get the 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}; apply it twice to get I4I_4.

Let In=∫sin⁡nx dx=∫sin⁡n−1x⋅sin⁡x dxI_n=\displaystyle\int\sin^nx\,dx=\int\sin^{n-1}x\cdot\sin x\,dx

Integrate by parts with u=sin⁡n−1xu=\sin^{n-1}x, dv=sin⁡x dxdv=\sin x\,dx, so du=(n−1)sin⁡n−2xcos⁡x dxdu=(n-1)\sin^{n-2}x\cos x\,dx, v=−cos⁡xv=-\cos x:

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

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

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

Deduce I4I_4: With n=4n=4:

I4=−sin⁡3xcos⁡x4+34I2I_4=-\dfrac{\sin^3x\cos x}{4}+\dfrac34I_2

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