First slide

Introduction to integration

Question

$\int \cos (\log_{e} x) dx is equal to$

Moderate

A

$\frac{1}{2} x [\cos (\log_{e} x) + \sin (\log_{e} x)]$
B

$x [\cos (\log_{e} x) + \sin (\log_{e} x)]$
C

$\frac{1}{2} x [\cos (\log_{e} x) - \sin (\log_{e} x)]$
D

$x [\cos (\log_{e} x) - \sin (\log_{e} x)]$

Solution

$\begin{array}{l} I = \int \cos (\log_{e} x) dx = \int \cos (\log_{e} x) \cdot 1 dx \\ = \cos (\log_{e} x) x - \int \frac{- \sin (\log_{e} x)}{x} \cdot xdx \\ = xcos (\log_{e} x) + \int \sin (\log_{e} x) dx \\ = xcos (\log_{e} x) + \int \sin (\log_{e} x) \cdot 1 dx \\ = xcos (\log_{e} x) + \sin (\log_{e} x) x - \int \frac{\cos (\log_{e} x)}{x} xdx \\ = xcos (\log_{e} x) + xsin (\log_{e} x) - I \\ 2 I = x [\cos (\log_{e} x) + \sin (\log_{e} x)] \\ I = \frac{x}{2} [\cos (\log_{e} x) + \sin (\log_{e} x)] \end{array}$

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