求下列不定积分。。。。。。。。。。求详解
(1)∫[(cos2x)/(cosx﹢sinx)]dx;(2)∫cot²xdx;(3)∫{(1+2x²)/[x²(1+x²)]}d...
(1)∫[(cos 2x)/(cos x﹢sin x)]dx;
(2) ∫cot²xdx;
(3) ∫{(1+2x²)/[x²(1+x²)]}dx;
(4) ∫sin²(x/2)dx;
(5) ∫[(cos2x)/(sin²xcos²x)]dx;
(6) ∫[e^(x-4)]dx;
求解答啊。。。。。。。。。。。。。。。。谢谢啦 展开
(2) ∫cot²xdx;
(3) ∫{(1+2x²)/[x²(1+x²)]}dx;
(4) ∫sin²(x/2)dx;
(5) ∫[(cos2x)/(sin²xcos²x)]dx;
(6) ∫[e^(x-4)]dx;
求解答啊。。。。。。。。。。。。。。。。谢谢啦 展开
1个回答
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cos2x=cos²x-sin²x=(cosx+sinx)(cosx-sinx)
∫(cosx-sinx)dx=sinx+cosx+C
∫cot²xdx=∫(csc²x-1)dx=-cotx-x+C
x=tanθ,dx=sec²θdθ
∫(1+2tan²θ)sec²θdθ/tan²θsec²θ
=∫(cotθ+2)dθ=ln|sinθ|+2θ+C=ln|x/√(1+x²)|+2arctanx+C
∫sin²(x/2)dx=(1/2)∫(1-cosx)dx=(x-sinx)/2+C
∫cos2xdx/(sinxcosx)²=4∫cos2xdx/sin²2x
=2∫d(sin2x)/sin²2x=-2/sin2x+C
∫[e^(x-4)]dx=e^(x-4)+C
∫(cosx-sinx)dx=sinx+cosx+C
∫cot²xdx=∫(csc²x-1)dx=-cotx-x+C
x=tanθ,dx=sec²θdθ
∫(1+2tan²θ)sec²θdθ/tan²θsec²θ
=∫(cotθ+2)dθ=ln|sinθ|+2θ+C=ln|x/√(1+x²)|+2arctanx+C
∫sin²(x/2)dx=(1/2)∫(1-cosx)dx=(x-sinx)/2+C
∫cos2xdx/(sinxcosx)²=4∫cos2xdx/sin²2x
=2∫d(sin2x)/sin²2x=-2/sin2x+C
∫[e^(x-4)]dx=e^(x-4)+C
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