Answer to Question #186236 in Calculus for Phyroe

Question

Integration by Partial Fractions

∫dt/(1+sint+cost)

Answer 1

∫dt/(1 + sint + cost ) ......................(1)

Let tan t/2 = z

Differentiating the above equation with respect to t, we have

(1/2 ) sec² ( t/2 ) dt = dz

dt = 2 dz/( sec² ( t/2 ) )

dt = 2 dz/( 1 + z²) .....................[∵ sec² x = 1+ tan² x]

Again by half angle formulae we have

sin t = 2 tan ( t/2) /( 1+ tan² (t/2) )

cos t = ( 1- tan² (t/2) ) / ( 1+ tan² t/2) )

Substituting the values of sin t, cos t and dt from above in equation (1) we have

∫ [ 2 dz/ ( 1 + z²) ] / [ 1+ 2z/(1 + z²) +  [ ( 1- z²) / ( 1+ z²) ]

∫ [2 dz / ( 1 + z²) ] / [ (2 +2z) / ( 1 + z²) ]

∫ 2 dz /( 2+ 2z)

∫ dz /(1+z)

ln | 1+z| + c

Substituting the value of z from above we have

ln | 1+ tan (t/2) | + c