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  4. For n ge 2, if In = ∫ ( sin x + cos x)n dx then n In - 2(n - 1) In - 2 =

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Q. For $n \ge 2$, if $I_n = \int (\sin x + \cos x)^n dx$ then $n I_n - 2(n - 1) I_{n - 2} = $

AP EAMCETAP EAMCET 2019

Solution:

We have,
$I_{n}=\int(\sin x+\cos x)^{n} d x$
$=\int(\sin x+\cos x)^{n-1} \cdot(\sin x+\cos x) d x$
$=\int(\sin x+\cos x)^{n-1}(\sin x-\cos x)$
$-\int(n-1)(\sin x+\cos x)^{n-2}(\cos x-\sin x)(\sin x-\cos x) d x$
Integration by parts,
$\therefore I_{n}=(\sin x+\cos x)^{n-1}(\sin x-\cos x)$
$+(n-1) \int(\sin x+\cos x)^{n-2}\left\{(\sin x-\cos x)^{2}\right\} d x$
As we know that,
$(\sin x+\cos x)^{2}+(\sin x-\cos x)^{2}=2$
$I_{n}=(\sin x+\cos x)^{n-1}(\sin x-\cos x)+(n-1)$
$\int(\sin x+\cos x)^{n-2} 2-(\sin x+\cos x)^{2} d x$
$=(\sin x+\cos x)^{n-1}(\sin x-\cos x)+(n-1)$
$\int 2(\sin x+\cos x)^{n-2} d x-(n-1) \int(\sin x+\cos x) d x$
$=(\sin x+\cos x)^{n-1}(\sin x-\cos x)+2(n-1)$
$I_{n-2}=(n-1) I_{n}$
$\Rightarrow n I_{n}-2(n-1) I_{n-2}=(\sin x+\cos x)^{n-1}$
$(\sin x-\cos x)+c$