`int(dx)/(sinx-cosx+sqrt2)`は次式に等しい。
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微積分ヘルプ: (sinx-cosx)^2 dx の積分 - 三角関数の恒等式による積分
`int (sinx+cosx)/sqrt(1+sin2x)dx`.
Integral {sqrt(sinx)/[sqrt(sinx)+sqrt(cosx)]}dx from 0 to pi/2 ||
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😳 きれいな基本微分 d/dx(sin2x)=? を微分します。 #ショートパンツ
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`int_0^(pi/4)(sqrt(tanx))/(sinxcosx)dx`
Show that ∫(x→0,π/2)sin2x/(sinx + cosx) = (1/√2)log ( √2+1) Integration 12th Important Question
Show that 0 to pi/2 integral x/sinx+cosx dx = π/2√2 log (√2+1) || Vvvip || Definite Integration
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Evaluate `int(sqrt(tan x))/(sin x cos x)d x`
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integrate sin x + cos x upon root 1 + sin 2x