**True or false**:

By double angle identities, we know that \( \cos2x = 2\cos^2x - 1 = 1-2\sin^2 x\).

\[ \begin{eqnarray} \int 2\sin 2x \, dx &=& -\cos 2x + C \\ &=&- 2\cos^2x + 1 + C \\ &=& - 2\cos^2x + C \end{eqnarray} \]

\[ \begin{eqnarray} \int 2\sin 2x \, dx &=& -\cos 2x + C \\ &=& -1 + 2\sin^2 x + C \\ &=& 2\sin ^2x + C \end{eqnarray} \]

From the two indefinite integrals below, we can see that \[ - 2\cos^2x + C = 2\sin ^2x + C \; . \]

Cancelling off the arbitrary constant of integration \(C\), we obtain \(-2\cos^2 x = 2\sin^2x \) or equivalently \( \cos^2 x =- \sin^2 x \) is true for all \(x\).

×

Problem Loading...

Note Loading...

Set Loading...