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# Integration by trig substitution

B.Sudhakar
09/01/2017 0 0

Let's solve one more trig problem using trig substitution.

∫ 5/√(x² - 5 ) dx    as variable x precedes constant inside the radical sign we may have substitute x = √5 sec u

Let x =  √5 sec u    dx = √5 sec u tan u du

= ∫ 5•√5 sec u tan u du /√( 5 sec² u - 5 )

= ∫ 5•√5 sec u tan u du /√5 √( sec² u - 1 )

=∫ 5 • √5 sec u tan u du /√ 5 √( tan² u ).                      ( sec² u - 1 = tan² u )

= ∫ 5 sec u tan u du / tan u

= 5 ∫ sec u du

= 5 ln | sec u + tan u | + C.                                         ( substitute u = arcsec x/√5 )

= 5 ln | sec arcsec x/√5 + tan arcsec x/√5 | + C.

= 5 ln | x/√5 + (√x² - 5)/√5 | + C

= 5 ln | x/√5 +  √{(x²/5) - 1)} | + C

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