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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 2025Subjective· 7mImportance★★★★★
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Integrate In=∫sin⁡nx dxI_n=\int\sin^nx\,dx by parts, splitting off one factor of sin⁡x\sin x and using cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x to relate InI_n to In−2I_{n-2}.

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 (du=(n−1)sin⁡n−2xcos⁡x dxdu=(n-1)\sin^{n-2}x\cos x\,dx) and dv=sin⁡x dxdv=\sin x\,dx (v=−cos⁡xv=-\cos x):

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

Using cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x:

In=−sin⁡n−1xcos⁡x+(n−1)∫sin⁡n−2x dx−(n−1)∫sin⁡nx dx=−sin⁡n−1xcos⁡x+(n−1)In−2−(n−1)In.I_n=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}x\,dx-(n-1)\int\sin^nx\,dx = -\sin^{n-1}x\cos x+(n-1)I_{n-2}-(n-1)I_n.

Collecting InI_n terms:

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

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

Deducing ∫sin⁡4x dx\int\sin^4x\,dx (i.e. I4I_4):

I4=−sin⁡3xcos⁡x4+34I2.I_4=-\frac{\sin^3x\cos x}{4}+\frac34I_2.

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