 | nx=tと置換すると
∫[0→π/2](sinnx)^2/(1+x)dx=∑[k=1～n]∫[(k-1)π/2→kπ/2](sint)^2/(n+t)dt
また、(k-1)π/2≦t≦kπ/2においてはn+(k-1)π/2≦n+t≦n+kπ/2だから
∴(sint)^2/(n+kπ/2)≦(sint)^2/(n+t)≦(sint)^2/(n+(k-1)π/2)
∴∫[(k-1)π/2→kπ/2](sint)^2/(n+kπ/2)dt≦∫[(k-1)π/2→kπ/2](sint)^2/(n+t)dt≦∫[(k-1)π/2→kπ/2](sint)^2/(n+(k-1)π/2)dt
∴(π/2)/(2n+kπ)≦∫[(k-1)π/2→kπ/2](sint)^2/(n+t)dt≦(π/2)/(2n+(k-1)π)
∴∑[k=1～n](π/2)/(2n+kπ)≦∑[k=1～n]∫[(k-1)π/2→kπ/2](sint)^2/(n+t)dt≦∑[k=1～n](π/2)/(2n+(k-1)π)
この式の最左辺＆最右辺はn→∞のとき(1/2)log(π/2+1)
よって挟み撃ちの原理により∑[k=1～n]∫[(k-1)π/2→kπ/2](sint)^2/(n+t)dt→(1/2)log(π/2+1) (n→∞)
以上よりlim[n→∞]∫[0→π/2](sinnx)^2/(1+x)dx=(1/2)log(π/2+1)
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