Question

求大神解高数_(´ཀ`」 ∠)__

Answer 1

设x=3/cosα 则：dx=3sinα/cos²αdα
∫√(x²-9)³dx
=∫√(9/cos²α-9)³dx
=27∫tan³α3sinα/cos²αdα
=81∫tan³αsinαd(tanα)
=(81/4)∫sinαd(tanα)^4
=(81/4)sinα(tanα)^4-(81/4)∫(tanα)^4dsinα
=(81/4)sinα(tanα)^4-(81/4)∫(tanα)^4cosαdα
=(81/4)sinα(tanα)^4-(81/4)∫(tanαsin³α)/cos²αdα
=(81/4)sinα(tanα)^4-(81/4)∫sin³αd(tanα)
=(81/4)sinα(tanα)^4-(81/4)sin³αtanα+(81/4)∫tanαdsin³α
=(81/4)sinα(tanα)^4-(81/4)sin³αtanα+(81/4)∫tanα3sin²αcosαdα
=(81/4)sinα(tanα)^4-(81/4)sin³αtanα+(243/4)∫sinα(1-cos²α)dα
=(81/4)sinα(tanα)^4-(81/4)sin³αtanα-(243/4)cosα+(81/4)cos³α+C