前>>563
>>589(1)
x=tanθとおくとx=sinθ/cosθだから、
dx/dθ=｛(sinθ)'cosθ-sinθ(cosθ)'｝/cos^2θ
=｛cos^2θ-(-sin^2θ)｝/cos^2θ
=1/cos^2θ
dx=dθ/cos^2θ
I(1)=∫[x=0→1]dx/(x^2+1)
=∫[θ=0→π/4]dθ/｛cos^2θ(tan^2θ+1)｝
=∫[θ=0→π/4]dθ
=[θ](θ=0→π/4)
=π/4