∫(7x2−3)32⋅dx
After using x=√37⋅secu and dx=√37⋅secu⋅tanu⋅du transforms, this integral became,
∫[7⋅37(secu)2−3]32⋅√37⋅secu⋅tanu⋅du
=∫[3(secu)2−3]32⋅√37⋅secu⋅tanu⋅du
=∫[3(tanu)2]32⋅√37⋅secu⋅tanu⋅du
=∫[3√3(tanu)3⋅√37⋅secu⋅tanu⋅du
=∫[9√7⋅(tanu)4⋅secu⋅du
Now, I solved ∫(tanu)4⋅secu⋅du integral,
∫(tanu)4⋅secu⋅du
=∫[(secu)2−1]2⋅secu⋅du
=∫(secu)5⋅du−2∫(secu)3⋅du+∫secu⋅du
After using reduction formula for ∫(secu)n⋅du,
∫(secu)n⋅du=1n−1⋅(secu)n−2⋅tanu+n−2n−1⋅∫(secu)n−2⋅du
∫(tanu)4⋅secu⋅du
=14⋅(secu)3⋅tanu+34∫(secu)3⋅du−2∫(secu)3⋅du+∫secu⋅du
=14⋅(secu)3⋅tanu−54∫(secu)3⋅du+∫secu⋅du
=14⋅(secu)3⋅tanu−58⋅secu⋅tanu−58∫secu⋅du+∫secu⋅du
=14⋅(secu)3⋅tanu−58⋅secu⋅tanu+38∫secu⋅du
=14⋅(secu)3⋅tanu−58⋅secu⋅tanu+38ln(secu+tanu)+√79⋅C1
Hence,
=∫9√7⋅(tanu)4⋅secu⋅du
=94√7⋅(secu)3⋅tanu-458√7⋅secu⋅tanu+278√7⋅ln(secu+tanu)+C1
After using x=√37⋅secu, secu=√73⋅x and tanu=√7x2−3√3 inverse transformations, I found
∫(7x2−3)32⋅dx
=94√7⋅7√79⋅x3⋅√7x2−3-458√7⋅√73⋅x√7x2−3+278√7⋅ln(√73⋅x+√7x2−3√3)+C1
=74x3⋅√7x2−3-158x√7x2−3+27√756⋅ln(x√7+√7x2−3)+C
Note: C=C1−27√7112⋅ln3
