6.12d

\(\frac{d}{dx}\left( \int_{\cos x}^{\sin x}{\sqrt{1-t^{2}}dt} \right),\; o<x<\frac{\pi }{2}\)

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Eftersom sinx och cosx ≥ 0 i definitionsmängden så kan integralen delas upp i komponenterna 

\(\int_{0}^{\sin x}{\sqrt{1-t^{2}}dt}-\int_{0}^{\cos x}{\sqrt{1-t^{2}}dt}\)

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\(f\left( x \right):=\int_{0}^{x}{\sqrt{1-t^{2}}dt}\)

Vilket enligt analysens huvudsats ger

\(f'\left( x \right)=\sqrt{1-x^{2}}\)

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Allt ovan ger följande tämligen utbroderade uträkning.

\(\frac{d}{dx}\left( \int_{\cos x}^{\sin x}{\sqrt{1-t^{2}}dt} \right)=\frac{d}{dx}\left( \int_{0}^{\sin x}{\sqrt{1-t^{2}}dt}-\int_{0}^{\cos x}{\sqrt{1-t^{2}}dt} \right)=\)

\(=\frac{d}{dx}f\left( \sin x \right)-\frac{d}{dx}f\left( \cos x \right)=f'\left( \sin x \right)\cdot \frac{d}{dx}\left( \sin x \right)-f'\left( \cos x \right)\cdot \frac{d}{dx}\left( \cos x \right)=\)

\(=\cos \left( x \right)\sqrt{1-\sin ^{2}x}-\left( -\sin x \right)\sqrt{1-\cos ^{2}x}=\cos \left( x \right)\sqrt{\cos ^{2}x}+\sin \left( x \right)\sqrt{\sin ^{2}x}=\)

\(=\cos ^{2}\left( x \right)+\sin ^{2}\left( x \right)=1\)