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式cos(A + B)= Cos A cos B-sin A sin B、 cos2xの式を取得します。
つまり、Cos 2x = Cos(x + x)= Cos x cos x –sin x sin x = cos ^ 2 x –sin ^ 2 x
つまり、 Cos 2x = cos ^ 2 x –sin ^ 2 x … ……。(1)
繰り返しますが、 cos ^ 2 x –sin ^ 2 x = cos ^ 2 x-(1- cos ^ 2 x) = 2 cos ^ 2 x-1
つまり、 Cos 2x = 2 cos ^ 2 x-1 ……………..(2)
繰り返しますが、2 cos ^ 2 x –1 = 2(1-sin ^ 2 x)-1 = 1-2 sin ^ 2 x
So 、Cos 2x = 1-2 sin ^ 2 x
したがって、 Cos 2x = cos ^ 2 x –sin ^ 2 x = 2 cos ^ 2 x-1 = 1-2 sin ^ 2 x
回答
これはab * tchです:
{\ displaystyle \ int} \ dfrac {2x} {\ sin \ left(2x \ right)} \、\ mathrm {d} x
この方程式を= {\ displaystyle \ int} 2x \ csc \ left(2x \ right)\、\ mathrm {d} x \ Rightarrow \ class {steps-node} {\ cssId {steps- node-1} {2}} {\ displaystyle \ int} x \ csc \ left(2x \ right)\、\ mathrm {d} x
{\ displaystyle \ int} xについて解きましょう\ csc \ left(2x \ right)\、\ mathrm {d} x
指数を使用した書き換え:= {\ displaystyle \ int} \ dfrac {2 \ mathrm {i} x} {\ mathrm { e} ^ {2 \ mathrm {i} x}-\ mathrm {e} ^ {-2 \ mathrm {i} x}} \、\ mathrm {d} x \ Rightarrow \ class {steps-node} {\ cssId {steps-node-2} {2 \ mathrm {i}}} {\ cdot} {\ displaystyle \ int} \ dfrac {x} {\ mathrm {e} ^ {2 \ mathrm {i} x}-\ mathrm {e} ^ {-2 \ mathrm {i} x}} \、\ mathrm {d} x
解決:{\ displaystyle \ int} \ dfrac {x} {\ mathrm {e} ^ {2 \ mathrm {i} x}-\ mathrm {e} ^ {-2 \ mathrm {i} x}} \、\ mathrm {d} x
共通の分母に用語を置きます: = {\ displaystyle \ int} \ dfrac {x \ mathrm {e} ^ {2 \ mathrm {i} x}} {\ mathrm {e} ^ {4 \ mathrm {i} x} -1} \、\ mathrm {d} x
\ dfrac {\ mathrm {d}} {\ mathrm {d} x} \ left [\ mathrm {i} x \ right] = \ mathrm {i}。mathtt {g}
\ mathtt {f} \:= \ ln \ left(v \ right)\ to \ mathtt {f} “\:= \ dfrac {1} {v}; \ mathtt {g} “\:= \ dfrac {1} {v + 1} \ to \ mathtt {g} \:= \ ln \ left(v + 1 \ right)
= \ ln \ left (v \ right)\ ln \ left(v + 1 \ right)-{\ displaystyle \ int} \ dfrac {\ ln \ left(v + 1 \ right)} {v} \、\ mathrm {d} v
{\ displaystyle \ int} \ dfrac {\ ln \ left(v + 1 \ right)} {v} \、\ mathrm {d} v
解決:{\ displaystyle \ int} \ dfrac {\ ln \ left(v + 1 \ right)} {v} \、\ mathrm {d} v
w = -v \ longrightarrow \ mathrm {d} vに置き換えます=-\ mathrm {d} w
=-{\ displaystyle \ int}-\ dfrac {\ ln \ left(1-w \ right)} {w} \、\ mathrm {d} w
これも上記のような対数です:= \ operatorname {Li} \_2 \ left(w \ right)
つまり、-{\ displaystyle \ int}-\ dfrac {\ ln \ left(1-w \ right)} {w} \、\ mathrm {d} w =-\ operatorname {Li} \_2 \ left(w \ right)
置換を元に戻すw = -v: =-\ operatorname {Li} \_2 \ left(-v \ right)
解決された積分をプラグインします:\ ln \ left(v \ right)\ ln \ left(v + 1 \ right)-{ \ displaystyle \ int} \ dfrac {\ ln \ left(v + 1 \ right)} {v} \、\ mathrm {d} v = \ ln \ left(v \ right)\ ln \ left(v + 1 \右)+ \ operatorname {Li } \_2 \ left(-v \ right)
次の解を求めます:{\ displaystyle \ int} \ dfrac {\ ln \ left(v \ right)} {v-1} \、\ mathrm { d} v
w = v-1 \ longrightarrow \ mathrm {d} v = \ mathrm {d} w
= {\ displaystyle \ int} \ dfrac {\ lnに置き換えます\ left(w + 1 \ right)} {w} \、\ mathrm {d} w
繰り返しますが、これは対数です:=-\ operatorname {Li} \_2 \ left(-w \ right )
そして、w = v-1なので:=-\ operatorname {Li} \_2 \ left(1-v \ right)
解決された積分をプラグインします:-\ class {steps -node} {\ cssId {steps-node-8} {\ dfrac {1} {2}}} {\ displaystyle \ int} \ dfrac {v \ ln \ left(v \ right)} {v ^ 2 + 1 } \、\ mathrm {d} v + \ class {steps-node} {\ cssId {steps-node-9} {\ dfrac {1} {4}}} {\ displaystyle \ int} \ dfrac {\ ln \ left (v \ right)} {v + 1} \、\ mathrm {d} v + \ class {steps-node} {\ cssId {steps-node-10} {\ dfrac {1} {4}}} {\ displaystyle \ int} \ dfrac {\ ln \ left(v \ right)} {v-1} \、\ mathrm {d} v =-\ dfrac {\ ln \ left(v \ right)\ ln \ left(v ^ 2 + 1 \ right)} {4}-\ dfrac {\ operatorname {Li} \_2 \ left(-v ^ 2 \ right)} {8} + \ dfrac {\ ln \ left(v \ right)\ ln \ left(v + 1 \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left(-v \ right)} {4}-\ dfrac {\ operatorname {Li} \_2 \ left(1-v \正しい)} {4}
v = \ mathrm {e} ^ uの置換を元に戻し、\ ln \ left(\ mathrm {e} ^ u \ right)= uを使用します:
=-\ dfrac {u \ ln \ left(\ mathrm {e} ^ {2u} +1 \ right)} {4}-\ dfrac {\ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {2u} \ right)} {8} + \ dfrac {u \ ln \ left(\ mathrm {e} ^ u + 1 \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left(- \ mathrm {e} ^ u \ right)} {4}-\ dfrac {\ operatorname {Li} \_2 \ left(1- \ mathrm {e} ^ u \ right)} {4}
解いた積分を再度プラグインします:-{\ displaystyle \ int} \ dfrac {u \ mathrm {e} ^ {2u}} {\ mathrm {e} ^ {4u} -1} \、\ mathrm {d} u = \ dfrac {u \ ln \ left(\ mathrm {e} ^ {2u} +1 \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {2u} \右)} {8}-\ dfrac {u \ ln \ left(\ mathrm {e} ^ u + 1 \ right)} {4}-\ dfrac {\ operatorname {Li} \_2 \ left(-\ mathrm {e } ^ u \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left(1- \ mathrm {e} ^ u \ right)} {4}
u = \なのでmathrm {i} x:= \ dfrac {\ mathrm {i} x \ ln \ left(\ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {2 \ mathrm {i} x} \ right)} {8}-\ dfrac {\ mathrm {i} x \ ln \ left(\ mathrm {e } ^ {\ mathrm {i} x} +1 \ right)} {4}-\ dfrac {\ operatorname {Li} \_2 \ l eft(-\ mathrm {e} ^ {\ mathrm {i} x} \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left(1- \ mathrm {e} ^ {\ mathrm {i } x} \ right)} {4}
解決された積分をプラグインします:\ class {steps-node} {\ cssId {steps-node-12} {2 \ mathrm {i}}} {\ cdot} {\ displaystyle \ int} \ dfrac {x} {\ mathrm {e} ^ {2 \ mathrm {i} x}-\ mathrm {e} ^ {-2 \ mathrm {i} x}} \、\ mathrm {d} x =-\ dfrac {x \ ln \ left(\ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right)} {2} + \ dfrac {\ mathrm {i} \ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {2 \ mathrm {i} x} \ right)} {4} + \ dfrac {x \ ln \ left(\ mathrm {e} ^ {\ mathrm {i} x} +1 \ right)} {2}-\ dfrac {\ mathrm {i} \ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {\ mathrm {i} x} \ right) } {2} + \ dfrac {\ mathrm {i} \ operatorname {Li} \_2 \ left(1- \ mathrm {e} ^ {\ mathrm {i} x} \ right)} {2}
解決された積分をプラグインします:\ class {steps-node} {\ cssId {steps-node-13} {2}} {\ displaystyle \ int} x \ csc \ left(2x \ right)\、\ mathrm {d } x = -x \ ln \ left(\ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right)+ \ dfrac {\ mathrm {i} \ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {2 \ mathrm {i} x} \ right)} {2} + x \ ln \ left(\ mathrm {e} ^ {\ mathrm {i} x} +1 \ right)-\ mathrm {i} \ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ { \ mathrm {i} x} \ right)+ \ mathrm {i} \ operatorname {Li} \_2 \ left(1- \ mathrm {e} ^ {\ mathrm {i} x} \ right)
ここで行う必要があるのは、不定積分のドメインを拡張するために、対数関数の引数に絶対値関数を適用することだけです。
{\ displaystyle \ int} 2x \ csc \ left(2x \ right )\、\ mathrm {d} x = -x \ ln \ left(\ left | \ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right | \ right)+ x \ ln \ left( \ left | \ mathrm {e} ^ {\ mathrm {i} x} +1 \ right | \ right)+ \ dfrac {\ mathrm {i} \ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {2 \ mathrm {i} x} \ right)} {2}-\ mathrm {i} \ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {\ mathrm {i} x} \ right)+ \ mathrm {i} \ operatorname {Li} \_2 \ left(1- \ mathrm {e} ^ {\ mathrm {i} x} \ right)+ C \ Rightarrow \ boxed {\ dfrac {\ mathrm {i} \ left (\ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {2 \ mathrm {i} x} \ right)-2 \ operatorname {Li} \_2 \ left(-\ mathrm {e} ^ {\ mathrm {i} x} \ right)+2 \ operatorname {Li} \_2 \ left(1- \ mathrm {e} ^ {\ mathrm {i} x} \ right)\ right)} {2} -x \ left( \ ln \ left(\ left | \ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right | \ right)-\ ln \ left(\ left | \ mathrm {e} ^ {\ mathrm { i} x} + 1 \ right | \ right)\ right)+ C}