不定积分习题 要详细过程 拜托了 快啊!!!
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∫(sinx+cosx)dx/(sinx-cosx)^3
=∫d(sinx-cosx)/(sinx-cosx)^3
=(-1/2)(1/(sinx-cosx)^2 +C
∫银神sinxdx/[(cosx)^3(1+(secx)^2)^(1/3)]
=-∫dcosx/[cosx^3(1+secx^2)^(1/3)]
=(1/2)∫dsecx^2/(1+secx^2)^(1/3)
=(1/2)(3/2)(secx^2+1)^(2/3) +C
∫dx/[x^2√(x^2+3)]
x=√3tanu dx=√3dtanu=√3secu^2du tanu=x/√晌历3 cotu=√3/x
1/sinu^2=cotu^2+1=[(3+x^2)/x^2 1/锋谨亏sinu=√(3+x^2)/x
=∫√3secu^2du/[tanu^2√3secu}
=∫cosudu/sinu^2
=-∫dsinu/sinu^2
=1/sinu+C
=√(3+x^2)/x+C
=
=∫d(sinx-cosx)/(sinx-cosx)^3
=(-1/2)(1/(sinx-cosx)^2 +C
∫银神sinxdx/[(cosx)^3(1+(secx)^2)^(1/3)]
=-∫dcosx/[cosx^3(1+secx^2)^(1/3)]
=(1/2)∫dsecx^2/(1+secx^2)^(1/3)
=(1/2)(3/2)(secx^2+1)^(2/3) +C
∫dx/[x^2√(x^2+3)]
x=√3tanu dx=√3dtanu=√3secu^2du tanu=x/√晌历3 cotu=√3/x
1/sinu^2=cotu^2+1=[(3+x^2)/x^2 1/锋谨亏sinu=√(3+x^2)/x
=∫√3secu^2du/[tanu^2√3secu}
=∫cosudu/sinu^2
=-∫dsinu/sinu^2
=1/sinu+C
=√(3+x^2)/x+C
=
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