∫sin(ln(x))dx=?


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Question Number 156695 by amin96 last updated on 14/Oct/21

$\int s i n (l n (x)) d x = ?$
	Answered by puissant last updated on 10/Nov/21
	$Q=∫_0 ^1 sin(lnx)dx ; u=lnx→x=e^u →dx=e^u du ⇒ Q=∫_0 ^1 e^u sinu du IBP : { ((i=sinu)),((j′=e^u )) :} ⇒ { ((i′=cosu)),((j=e^u )) :} ⇒ Q = [e^u sinu]_0 ^1 −∫_0 ^1 e^u cosu du Double IBP : { ((i=cosu)),((j′=e^u )) :} ⇒ { ((i′=−sinu)),((j=e^u )) :} ⇒ Q= [e^u sinu−e^u cosu]_0 ^1 −∫_0 ^1 e^u sinu du ⇒ Q ={ (e^u /2)(sinu−cosu)}_0 ^1 ⇒ Q = {(e/2)(sin(1)−cos(1))}+(1/2) ∴∵ Q = (1/2){e(sin(1)−cos(1))+1}.. ..................Le puissant.................$
	$Q = \int_{0}^{1} s i n (l n x) d x; u = l n x \to x = e^{u} \to d x = e^{u} d u$ $\Rightarrow Q = \int_{0}^{1} e^{u} s i n u d u$ $I B P : {\begin{cases} i = s i n u \\ j^{'} = e^{u} \end{cases} \Rightarrow {\begin{cases} i^{'} = c o s u \\ j = e^{u} \end{cases}$ $\Rightarrow Q = {[e^{u} s i n u]}_{0}^{1} - \int_{0}^{1} e^{u} c o s u d u$ $D o u b l e I B P : {\begin{cases} i = c o s u \\ j^{'} = e^{u} \end{cases} \Rightarrow {\begin{cases} i^{'} = - s i n u \\ j = e^{u} \end{cases}$ $\Rightarrow Q = {[e^{u} s i n u - e^{u} c o s u]}_{0}^{1} - \int_{0}^{1} e^{u} s i n u d u$ $\Rightarrow Q = {\frac{e^{u}}{2} (s i n u - c o s u)}_{0}^{1}$ $\Rightarrow Q = {\frac{e}{2} (s i n (1) - c o s (1))} + \frac{1}{2}$ $∴∵ Q = \frac{1}{2} {e (s i n (1) - c o s (1)) + 1} . .$ $. . . . . . . . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . . . . . . .$
	Commented by amin96 last updated on 14/Oct/21

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