integration

Question

show that ∫dx/(2+3sinx) [upper limit pi/2, lower limit 0] = ∫dt/((t+ 3/2)² - (√5/2)²) [ upper limit 1, lower limit 0]

 

 

I know that sin(x) = 2t/(t²+1), where t = tan(x/2) but i cannot get the answer they are asking for in the question. Help would be appreciated.

Answer 1

Solve the transform for x and differentiate:

 

t = tan(x/2)

 

x = 2 tan^-1 t

 

dx = 2/(1+t²) dt

 

We also must transform the bounds:

 

t_up = tan(x_up/2) = tan(π/4) = 1

 

t_low = tan(x_low/2) = tan 0 = 0

 

Now you correctly rewrote the transform, so this is familiar:

 

t² = tan²(x/2) = sec²(x/2) - 1

 

2/(1+t²) = 2cos²(x/2) = 1 + cos x

 

(1 - t²)/(1 + t²) = cos x

 

sin x = √(1 - cos² x) =  2t / (1+t²)

 

So now we can substitute to finish.

 

∫₀^π/2 dx/(2+3 sin x)

 

= ∫₀¹ [2dt/(1+t²)] / [2+6t/(1+t²)]

 

= ∫₀¹ dt / [(1+t²)+3t]

 

= ∫₀¹ dt / (t² + 3t + 9/4 - 5/4)

 

= ∫₀¹ dt / [(t + 3/2)² - (√5/2)²]