Question

For x2 + na + 1, n eN (the set of natural numbers), the integral 2sin(x-1)-sin 2(x² -1) dx "V2sin(x² 1) + sin 2( x² –1) is equal to : (where e is a constant of integration)

Answer 1

Answer:

your answer attacted in the photo

Answer 2

Answer:

${\huge{\boxed{\red{\mathscr{SolutioN}}}}}$

Let ,

$I \: = ∫x \: \sqrt{ \frac{2sin \: (x {}^{2} - 1) - sin {}^{2} (x {}^{2} - 1)}{2sin \: (x {}^{2} - 1) + sin {}^{2} (x {}^{2} - 1)} } \: \: dx$

Put ,

$x {}^{2} - 1 = t \\ = > \: 2x \: = \frac{dt}{dx} \\ x \: dx \: \: = \frac{dt}{2}$

Then ,

$I \: = ∫ \sqrt{ \frac{2sin \: t \: - 2sin \: t}{2sin \: t \: + 2sin \: t} } \: \: \: \frac{dt}{2}$

$= \frac{1}{2} ∫ \sqrt{ \frac{2sin \: t \: - 2sin \: t \: cos \: t}{2sin \: t \: + \: 2sin \: t \: - cos \: t} } \: \: dt$

$= \frac{1}{2} ∫ \sqrt{ \frac{1 - cos \: t}{1 \: + \: cos \: t} } \: \: dt$

$= \frac{1}{2} ∫ \sqrt{ \frac{2sin {}^{2} \frac{t}{2} }{2cos {}^{2} \frac{t}{2} } } \: \: dt$

$= \frac{1}{2} ∫ \: tan \frac{t}{2} \: \: dt$

$= \frac{1}{2} \: \: \: \: \frac{log \: │sec \frac{t}{2} │}{ \frac{1}{2} } + c$

$log \: │sec \frac{t}{2} │ + c$

$= log \: │ \: sec \: ( \: \frac{x {}^{2} - 1}{2} ) \: │ + \: c$