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If $\displaystyle I_{k}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^ndx\;,n\in \mathbb{N}$ Then Relation between $I_{k}$ and $I_{k+2}$

$\bf{My\; Try::}$ Given $\displaystyle I_{k}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^ndx\;,$ Then $\displaystyle I_{k+2}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^{n+2}dx\;,$

So $$I_{k+2}-I_{k} = \int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^n\cdot \sin 2xdx$$

Now How can I solve it after that, Help me

Thanks

juantheron
  • 53,015

1 Answers1

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Note: $$\int_0^\frac{\pi}{2} x(\sin x+\cos x)^n dx=\frac{\pi}{4}\int_0^\frac{\pi}{2}(\sin x+\cos x)^ndx=I_n$$ (Put $x=\frac{\pi}{2}-x$ to prove this)

We have $$I_{n+2}-I_n=\frac{\pi}{4}\int_0^\frac{\pi}{2} (\sin x+\cos x)^n \sin 2xdx$$ Applying by parts, differentiating $(\sin x+\cos x)^n$ and integrating $\sin 2x$, this becomes: $$I_{n+2}-I_n=\frac{\pi}{4}\left (-\frac 12 \cos 2x(\sin x+\cos x)^n|_0^{\frac{\pi}{2}}+\frac{n}{2}\int_0^\frac{\pi}{2}(\sin x+\cos x)^n dx-\frac{n}{2}\int_0^\frac{\pi}{2}(\sin x+\cos x)^n \sin 2x dx\right)$$

Now use $$\int_0^\frac{\pi}{2}(\sin x+\cos x)^n dx=\frac{4I_n}{\pi}$$ and $$\int_0^\frac{\pi}{2}(\sin x+\cos x)^n \sin 2x dx=\frac{4(I_{n+2}-I_n)}{\pi}$$ to get: $$\color{green} {I_{n+2}=I_n+\frac{\pi}{2(n+2)}}$$

Nikunj
  • 6,160