Question

Integration of under root 1 + sin 2 theta​

Answer 1

`\sin \theta - \cos \theta + c`

Explanation:

`\int √(1 + \sin 2 \theta) \: d \theta \\ formulae \: to \: be \: ued \\ { \sin}^(2) \theta + { \cos}^(2) \theta = 1 \\ \sin2 \theta = 2 \sin \theta \: \cos \theta \\ \\ \therefore \: \int √(1 + \sin 2 \theta) \: d \theta \\ = \int \sqrt{ { \sin}^(2) \theta + { \cos}^(2) \theta + 2 \sin \theta \: \cos \theta} \: d \theta \\ = \int \sqrt{ {(\sin \theta \: + \cos \theta )}^(2) } \: d \theta \\ = \int( \sin \theta + \cos \theta) \: d \theta \\ = \int \sin \theta \: d \theta + \int \cos \theta \: d \theta \\ = - \cos \theta + \sin \theta + c \\ \huge \red{ \boxed{ = \sin \theta - \cos \theta + c}}`