If I_n=int(cos ^n x+sin ^n x) d x and I_n-fra | vedclass

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(B) Given $I_n = \int (\cos^n x + \sin^n x) dx$.We can write $I_n = \int \cos^{n-1} x \cos x dx + \int \sin^{n-1} x \sin x dx$.Using integration by pa

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$\cos ^{n-2} x-\sin ^{n-2} x$

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(B) Given $I_n = \int (\cos^n x + \sin^n x) dx$.We can write $I_n = \int \cos^{n-1} x \cos x dx + \int \sin^{n-1} x \sin x dx$.Using integration by parts for both integrals:For $\int \cos^{n-1} x \cos x dx$: Let $u = \cos^{n-1} x$,$dv = \cos x dx$. Then $du = (n-1) \cos^{n-2} x (-\sin x) dx$,$v = \sin x$.$\int \cos^{n-1} x \cos x dx = \cos^{n-1} x \sin x - \int (n-1) \cos^{n-2} x (-\sin x) \sin x dx = \cos^{n-1} x \sin x + (n-1) \int \cos^{n-2} x \sin^2 x dx$.Since $\sin^2 x = 1 - \cos^2 x$,this becomes $\cos^{n-1} x \sin x + (n-1) \int \cos^{n-2} x (1 - \cos^2 x) dx = \cos^{n-1} x \sin x + (n-1) I_{n-2} - (n-1) I_n$.Similarly,for $\int \sin^{n-1} x \sin x dx$: Let $u = \sin^{n-1} x$,$dv = \sin x dx$. Then $du = (n-1) \sin^{n-2} x \cos x dx$,$v = -\cos x$.$\int \sin^{n-1} x \sin x dx = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \cos^2 x dx = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x (1 - \sin^2 x) dx = -\sin^{n-1} x \cos x + (n-1) I_{n-2} - (n-1) I_n$.Adding these results: $I_n = \cos^{n-1} x \sin x - \sin^{n-1} x \cos x + 2(n-1) I_{n-2} - 2(n-1) I_n$.Wait,the standard reduction formula for $I_n = \int \cos^n x dx$ is $I_n = \frac{\cos^{n-1} x \sin x}{n} + \frac{n-1}{n} I_{n-2}$.Applying this to both parts: $I_n = \left(\frac{\cos^{n-1} x \sin x}{n} + \frac{n-1}{n} I_{n-2}\right) + \left(\frac{-\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} I_{n-2}\right)$.This simplifies to $I_n = \frac{\cos^{n-1} x \sin x - \sin^{n-1} x \cos x}{n} + \frac{2(n-1)}{n} I_{n-2}$.However,the question implies $I_n - \frac{n-1}{n} I_{n-2}$. Let's re-evaluate: $I_n = \int \cos^n x dx + \int \sin^n x dx$.Using the reduction formula $I_n = \frac{\cos^{n-1} x \sin x}{n} + \frac{n-1}{n} I_{n-2}$ for $\cos^n x$ and $J_n = \int \sin^n x dx = \frac{-\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} J_{n-2}$.Then $I_n + J_n = \frac{\cos^{n-1} x \sin x - \sin^{n-1} x \cos x}{n} + \frac{n-1}{n} (I_{n-2} + J_{n-2})$.Thus $I_n - \frac{n-1}{n} I_{n-2} = \frac{\cos^{n-1} x \sin x - \sin^{n-1} x \cos x}{n} = \frac{\sin x \cos x}{n} (\cos^{n-2} x - \sin^{n-2} x)$.Therefore,$f(x) = \cos^{n-2} x - \sin^{n-2} x$.

If $I_n=\int(\cos ^n x+\sin ^n x) d x$ and $I_n-\frac{n-1}{n} I_{n-2}=\frac{\sin x \cos x}{n} f(x)$,then $f(x)=$

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