find-1-e-sin-ln-x-dx-

Question Number 32305 by abdo imad last updated on 22/Mar/18

f i n d \int_{1}^{e} s i n (l n (x)) d x .

Commented by abdo imad last updated on 24/Mar/18

l e t u s e t h e c h . l n x = t \Rightarrow I = \int_{0}^{1} s i n t e^{t} d t a n d

I = I m (\int_{0}^{1} e^{i t + t} d t) = I m (\int_{0}^{1} e^{(1 + i) t} d t) b u t

\int_{0}^{1} e^{(1 + i) t} d t = \frac{1}{1 + i} {[e^{(1 + i) t}]}_{0}^{1} = \frac{1}{1 + i} (e^{1 + i} - 1)

= \frac{1 - i}{2} (e (c o s (1) + i s i n (1) - 1)

= \frac{e c o s (1) + i e s i n (1) - 1 - i e c o s (1) + s i n (1) + i}{2}

= \frac{e c o s (1) + s i n (1) - 1 + i (e s i n (1) - e c o s (1) + 1)}{2}

I = \frac{1}{2} (e s i n (1) - e c o s (1) + 1) .

Answered by sma3l2996 last updated on 23/Mar/18

u = s i n (l n x) \Rightarrow u^{'} = \frac{1}{x} c o s (l n x)

v^{'} = 1 \Rightarrow v = x

s o \int_{1}^{e} s i n (l n x) d x = {[x s i n (l n x)]}_{1}^{e} - \int_{1}^{e} c o s (l n x) d x

= e s i n (1) - \int_{1}^{e} c o s (l n x) d x

u = c o s (l n x) \Rightarrow u^{'} = - \frac{1}{x} s i n (l n x)

v^{'} = 1 \Rightarrow v = x

\int_{1}^{e} s i n (l n x) d x = e . s i n (1) - {[x c o s (l n x)]}_{1}^{e} - \int_{1}^{e} s i n (l n x) d x

2 \int_{1}^{e} s i n (l n x) d x = e . s i n (1) - e . c o s (1) + 1

\int_{1}^{e} s i n (l n x) d x = \frac{e}{2} (s i n 1 - c o s 1) + \frac{1}{2}

Commented by abdo imad last updated on 24/Mar/18

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