# Thread: Techniques of integration (2)

1. ## Techniques of integration (2)

Evaluate $\int \frac{x^2}{4x \sin x + (4 - x^2) \cos x} \ dx.$

Hint:

Spoiler:

divide top and botttom of the integrand by "something" and then make a clever trig substitution!

2. I just give the final answer and the important step :

The final answer is - arccosh| cosec { x + arctan [(4-x^2 )/ ( 4x )] } | + c

And the important step is

4x sinx + ( 4 - x^2 ) cosx =
sqrt [ (4x)^2 + (4-x^2)^2 ] * sin { x + arctan[ (4-x^2)/(4x) ] }
= sqrt [ x^4 + 8x^2 + 16 ] * sin { x + arctan[ (4-x^2)/(4x) ] }
= ( x^2 + 4 ) * sin { x + arctan[ (4-x^2)/(4x) ] }

3. Originally Posted by simplependulum

$4x \sin x + ( 4 - x^2 ) \cos x = \sqrt{ (4x)^2 + (4-x^2)^2 } \ \sin \{ x + \arctan [(4-x^2)/(4x) ] \}$

$= \sqrt{x^4 + 8x^2 + 16} \ \sin \{ x + \arctan[ (4-x^2)/(4x) ] \}$

$= ( x^2 + 4 ) \sin \{ x + \arctan[ (4-x^2)/(4x) ] \}$
correct! then the substitution $x + \arctan[ (4-x^2)/(4x) ]=t$ will reduce the integral to the simple integral $\int \csc t \ dt.$