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Question: If $I_{m,n} = \int cos^m x cos nx dx$, then show that: $I_{m,n} = \frac{cos^m x sin nx}{m+n} + \frac...

If Im,n=∫cosmxcosnxdxI_{m,n} = \int cos^m x cos nx dx, then show that: Im,n=cosmxsinnxm+n+mm+nIm−1,n−1I_{m,n} = \frac{cos^m x sin nx}{m+n} + \frac{m}{m+n}I_{m-1, n-1}

Answer

cosmxsinnxm+n+mm+nIm−1,n−1\frac{cos^m x sin nx}{m+n} + \frac{m}{m+n}I_{m-1, n-1}

Explanation

Solution

To derive the reduction formula for Im,n=∫cos⁡mxcos⁡nx dxI_{m,n} = \int \cos^m x \cos nx \, dx, we use integration by parts.

Let u=cos⁡mxu = \cos^m x and dv=cos⁡nx dxdv = \cos nx \, dx. Then, we find dudu and vv: du=mcos⁡m−1x(−sin⁡x) dx=−mcos⁡m−1xsin⁡x dxdu = m \cos^{m-1} x (-\sin x) \, dx = -m \cos^{m-1} x \sin x \, dx v=∫cos⁡nx dx=sin⁡nxnv = \int \cos nx \, dx = \frac{\sin nx}{n}

Applying the integration by parts formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du: Im,n=cos⁡mx(sin⁡nxn)−∫(sin⁡nxn)(−mcos⁡m−1xsin⁡x) dxI_{m,n} = \cos^m x \left(\frac{\sin nx}{n}\right) - \int \left(\frac{\sin nx}{n}\right) (-m \cos^{m-1} x \sin x) \, dx Im,n=cos⁡mxsin⁡nxn+mn∫cos⁡m−1xsin⁡xsin⁡nx dxI_{m,n} = \frac{\cos^m x \sin nx}{n} + \frac{m}{n} \int \cos^{m-1} x \sin x \sin nx \, dx

Now, we need to simplify the integral term ∫cos⁡m−1xsin⁡xsin⁡nx dx\int \cos^{m-1} x \sin x \sin nx \, dx. We use the trigonometric identity: cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A-B) = \cos A \cos B + \sin A \sin B. Let A=nxA = nx and B=xB = x. Then cos⁡(nx−x)=cos⁡nxcos⁡x+sin⁡nxsin⁡x\cos(nx-x) = \cos nx \cos x + \sin nx \sin x. Rearranging this to find sin⁡nxsin⁡x\sin nx \sin x: sin⁡nxsin⁡x=cos⁡(n−1)x−cos⁡nxcos⁡x\sin nx \sin x = \cos(n-1)x - \cos nx \cos x

Substitute this identity into the integral: ∫cos⁡m−1xsin⁡xsin⁡nx dx=∫cos⁡m−1x(cos⁡(n−1)x−cos⁡nxcos⁡x) dx\int \cos^{m-1} x \sin x \sin nx \, dx = \int \cos^{m-1} x (\cos(n-1)x - \cos nx \cos x) \, dx =∫cos⁡m−1xcos⁡(n−1)x dx−∫cos⁡m−1xcos⁡nxcos⁡x dx= \int \cos^{m-1} x \cos(n-1)x \, dx - \int \cos^{m-1} x \cos nx \cos x \, dx The first integral is Im−1,n−1I_{m-1, n-1}. The second integral is ∫cos⁡mxcos⁡nx dx\int \cos^m x \cos nx \, dx, which is Im,nI_{m,n}. So, ∫cos⁡m−1xsin⁡xsin⁡nx dx=Im−1,n−1−Im,n\int \cos^{m-1} x \sin x \sin nx \, dx = I_{m-1, n-1} - I_{m,n}.

Substitute this back into the expression for Im,nI_{m,n}: Im,n=cos⁡mxsin⁡nxn+mn(Im−1,n−1−Im,n)I_{m,n} = \frac{\cos^m x \sin nx}{n} + \frac{m}{n} (I_{m-1, n-1} - I_{m,n}) Im,n=cos⁡mxsin⁡nxn+mnIm−1,n−1−mnIm,nI_{m,n} = \frac{\cos^m x \sin nx}{n} + \frac{m}{n} I_{m-1, n-1} - \frac{m}{n} I_{m,n}

Now, collect the Im,nI_{m,n} terms on the left side: Im,n+mnIm,n=cos⁡mxsin⁡nxn+mnIm−1,n−1I_{m,n} + \frac{m}{n} I_{m,n} = \frac{\cos^m x \sin nx}{n} + \frac{m}{n} I_{m-1, n-1} Im,n(1+mn)=cos⁡mxsin⁡nxn+mnIm−1,n−1I_{m,n} \left(1 + \frac{m}{n}\right) = \frac{\cos^m x \sin nx}{n} + \frac{m}{n} I_{m-1, n-1} Im,n(n+mn)=cos⁡mxsin⁡nxn+mnIm−1,n−1I_{m,n} \left(\frac{n+m}{n}\right) = \frac{\cos^m x \sin nx}{n} + \frac{m}{n} I_{m-1, n-1}

Finally, multiply both sides by nm+n\frac{n}{m+n} to isolate Im,nI_{m,n}: Im,n=nm+n(cos⁡mxsin⁡nxn+mnIm−1,n−1)I_{m,n} = \frac{n}{m+n} \left( \frac{\cos^m x \sin nx}{n} + \frac{m}{n} I_{m-1, n-1} \right) Im,n=cos⁡mxsin⁡nxm+n+mm+nIm−1,n−1I_{m,n} = \frac{\cos^m x \sin nx}{m+n} + \frac{m}{m+n} I_{m-1, n-1}

This completes the proof of the reduction formula.