求根下（x^2-a^2)的积分
1个回答
设x/a=sect,则dx=asect*tantdt,sint=√(x²-a²)/x
于是,原式=∫atant*asect*tantdt
=a²∫tan²t*sectdt
=a²∫sin²td(sint)/(1-sin²t)²
=a²/4∫[1/(1+sint)²+1/(1-sint)²-1/(1-sint)-1/(1+sint)]d(sint)
=a²/4[-1/(1+sint)+1/(1-sint)+ln│1-sint│-ln│1+sint│]+C1 (C1是积分常数)
=a²/4[ln│(1-sint)/(1+sint)│+2sint/(1-sin²t)]+C1
=a²/2[ln│x-√(x²-a²)│-ln│a│+x√(x²-a²)/a²]+C1
=a²/2ln│x-√(x²-a²)│+x√(x²-a²)/2+C1-a²/2ln│a│
=a²/2ln│x-√(x²-a²)│+x√(x²-a²)/2+C (C=C1-a²/2ln│a│,是积分常数).
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