Posted by **Mina** on Wednesday, May 14, 2008 at 2:15pm.

Solve the first-order initial value problem using half angle formulas

{y'(x)= 1 + sin^2 x

y (0)= -7

- Calculus -
**Reiny**, Wednesday, May 14, 2008 at 3:04pm
use the identity

cos 2x = 1 - 2sin^2 x

and solving for sin^2 x

sin^2 x = 1/2 - (1/2)cos 2x

then y'(x)= 1 + sin^2 x

becomes

y'(x)= 1 + sin^2 x

becomes

y'(x)= 1 + 1/2 - (1/2)cos 2x

= 3/2 - (1/2)cos 2x

this is now easy to integrate

Don't forget to add the constant.

plug in the point (0,-7) to find that constant.

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