∫ln x×cos 2ln xdx


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Question Number 27073 by sorour87 last updated on 01/Jan/18

$\int \ln x \times \cos 2 \ln x d x$
	Answered by prakash jain last updated on 02/Jan/18

	$u = \ln x$ $d u = \frac{1}{x} d x \Rightarrow d x = x d u \Rightarrow d x = e^{u} d u$ $= \int {u e}^{u} \cos 2 u d u$ $\cos 2 u = \frac{e^{i 2 u} + e^{- i 2 u}}{2} (i = \sqrt{- 1})$ $= \int u \frac{e^{(1 + 2 i) u} + e^{(1 - 2 i) u}}{2} d u$ $\int {u e}^{(1 + 2 i) u} d u$ $u \frac{e^{(1 + 2 i) u}}{1 + 2 i} - \int \frac{e^{(1 + 2 i) u}}{1 + 2 i} d u$ $u \frac{e^{(1 + 2 i) u}}{1 + 2 i} - \frac{e^{(1 + 2 i) u}}{{(1 + 2 i)}^{2}}$ $t o t a l i n t e g r a l$ $\frac{1}{2} [u \frac{e^{(1 + 2 i) u}}{1 + 2 i} - \frac{e^{(1 + 2 i) u}}{{(1 + 2 i)}^{2}} + u \frac{e^{(1 - 2 i) u}}{1 - 2 i} - \frac{e^{(1 - 2 i) u}}{{(1 w - 2 i)}^{2}}]$ $please recheck answer$ $to get the answer in \cos / \sin$ $substitute e^{i θ} = \cos θ + i \sin θ$
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