Question

Solve this...fast........

Answer 1

Answer:

=> ln(1+sin²x) + c

where c is an arbitrary constant

Step-by-step explanation:

$\int \frac{ \sin(2x) }{1 + \sin {}^{2} (x) } dx \\ \\ \bf \: let \: (1 + { \sin }^{2} (x)) \: =\: z \\ \\ \underline{ \bf \: differentiating \: \: both \: \: sides \: \: we \: \: get} \\ \\ = = > \frac{d(1 + \sin {}^{2} (x) )}{dx} = \frac{dz}{dx} \\ \\ = > \: 2 \sin(x) \cos(x). dx = dz \\ \\ \bf \underline{ note = > 2sinxcosx = \sin(2x) } \\ \\ = > \sin(2x) dx = dz \\ \\ \bf \: now \: substituting \: it \: in \: the \: question \: we \: get \\ \\ = > \int \frac{ \sin(2x) }{1 + \sin {}^{2} (x) } \\ \\ = \int \: \frac{dz}{z} \\ \\ = ln(z) + c \\ \\ = \bf{ \underline{ ln(1 + \sin {}^{2} (x) ) \: + c }}$