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$$\int2\sin x\cos x\mathrm dx$$ $1$:Consider $2\sin x\cos x=\sin(2x)$ Then the integration is $-\frac12\cos(2x)$

$2$: Consider $2\sin x\cos x=\frac{\mathrm d}{\mathrm dx}\sin^2(x)$ Then the integration is $\sin^2(x)$

If it is true then $-\frac12\cos(2x)=\sin^2(x)$ then $-\frac12(1+2\sin^2(x))=\sin^2(x)$,$-\frac12=\sin^2{x}$, $-\frac12=1$

mrtaurho
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alfatih Abdullah
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1 Answers1

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You are dealing with indefinite integral therefore you have to take into account that while integrating we have to add a constant of integration. To be precise we got that

\begin{align*} \int 2\sin x\cos x\mathrm dx&=\int 2\sin x\cos x\mathrm dx\\ \int \sin(2x)\mathrm dx&=\int \left(\frac{\mathrm d}{\mathrm dx}\sin^2(x)\right)\mathrm dx\\ -\frac12\cos(2x)+c_1&=\sin^2(x)+c_2\\ -\frac12\cos(2x)&=\sin^2(x)+c \end{align*}

To determine the constant of integration in this case we can follow your own attempt by expanding the $\cos(2x)$, even though you have made a minor mistake there, term to get

\begin{align*} -\frac12\cos(2x)&=\sin^2(x)+c\\ -\frac12(1-2\sin^2(x))&=\sin^2(x)+c\\ c&=-\frac12 \end{align*}

We just proved the Half-Angle Formula since we can now conclude that

$$\sin^2(x)=\frac12-\frac12\cos(2x)=\frac{1-\cos(2x)}2$$

Which is basically all what was missing.

mrtaurho
  • 16,103