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Require assistance with two rather complicated integration problems

 

Hi Mr Koh,

I am struggling with these two integrals; have spent more than an hour on them yet still couldn’t crack them apart. Please help!

∫sec^6 x dx and ∫(x-1)^6*(x-3)^2 dx

Student X

They do require a little ingenuity on your part; I have worked them out in full as follows:

∫sec^6 x dx = ∫sec^4 x sec^2 x dx

= ∫ (1+ tan^2 x)^2 * sec^2 x dx

= ∫ (1+ 2tan^2 x+tan^4 x) * sec^2 x dx

= ∫ sec^2 x + 2tan^2 x sec^2 x + tan^4 x sec^2 x dx

= tanx + 2/3*tan^3 x + 1/5 tan^5 x +C (shown)

∫(x-1)^6*(x-3)^2 dx = ∫(x-1)^6* [(x-1)-2]^2 dx

= ∫(x-1)^6* [(x-1)^2 -4(x-1) +4] dx

= ∫(x-1)^8 - 4(x-1)^7 + 4(x-1)^6 dx

= 1/9*(x-1)^9 - 1/2 *(x-1)^8 + 4/7*(x-1)^7 +C (shown)

Hope I have helped clarify matters. Peace.

Best Regards,

Mr Koh