Question

求几个定积分

你好，哪位高手帮忙求一下子子下面几个的积分。谢谢！

Answer 1

呵呵，分太少了吧

1.不太会。
原积分=[∫（0，π）+∫（π，2π）]√(1+cos²x)dx
经换元，得=2∫（0，π)√(1+cos²x)dx
然后就不会了

2.令u=cosx,则
=∫(1,-1)√（1+u^2）du
=-2∫(0,1)√（1+u^2）du
令u=tant,则du=sec²tdt
=-2∫(0,π/4)sec³tdt
用分部积分：
=-2∫（0,π/4）sectd(tant)
=-2[secttant(0,π/4)-∫（0,π/4）secttan²tdt]
=-2√2+2∫（0,π/4）sect(sec²t-1)dt
=-2√2+2∫(0,π/4)sec³tdt-2∫（0,π/4)sectdt
=-2√2+2∫(0,π/4)sec³tdt-2ln(sect+tant)(0,π/4)
=-2√2+2∫(0,π/4)sec³tdt-2ln(√2+1)
∴-2∫(0,π/4)sec³tdt=-2√2+2∫(0,π/4)sec³tdt-2ln(√2+1)
即-2∫(0,π/4)sec³tdt=√2+ln(√2+1)
即原积分=√2+ln(√2+1)

3.令x-10=3sint即x=10+3sint,则dx=3costdt
=∫(-π/2,π/2)(10+3sint)3cost*3costdt
=9∫(-π/2,π/2)(10+3sint)cos²tdt
=9[10∫(-π/2,π/2)cos²tdt+3∫(-π/2,π/2)sintcos²tdt]
=9[5∫(-π/2,π/2)(1+cos2t)dt-3∫(-π/2,π/2)cos²td(cost)]
=9[5(t+sin2t/2)(-π/2,π/2)-cos³t(-π/2,π/2)]
=45π

4.1/(1+x²)²=[(1+x²)/(1+x²)²+(1-x²)/(1+x²)²]/2
=1/2*[∫(1,2)(1+x²)/(1+x²)²dx+∫(1,2)(1-x²)/(1+x²)²dx]
=1/2*[∫(1,2)1/(1+x²)dx+∫(1,2)(1/x²-1)/(1/x+x)²dx]
=1/2[arctanx(1,2)-∫(1,2)(1-1/x²)/(1/x+x)²dx]
=1/2[arctan2-π/4)-∫(1,2)d(x+1/x)]/(1/x+x)²]
=1/2[arctan2-π/4+[1/(x+1/x)](1,2)]
=1/2[arctan2-π/4+(2/5-1/2)]
=(arctan2-π/4-1/10)/2

计算好多，但后三题方法肯定是对的，你算的时候再检验一下。