0Let'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
