Integration - Special Cases

Find   \( \int\ ln\ x\ dx \)

\begin{align} Let \quad I & = \int\ ln\ x\ dx \\ \\ & = \int\ (ln\ x \times 1)\ dx \\ \\ Using\ integration\ by\ parts \\ \\ Let\ u= ln\ x \\ \implies \frac{du}{dx} & = \frac{1}{x}\\ \\ Let\ \frac{dv}{dx}= 1 \\ \implies v & = x \\ \\ I & = x\ln\ x - \int\ \left(x \times \frac{1}{x} \right)\ dx \\ \\ x\ ln\ x - \int\ 1\ dx \\ \\ x\ln\ x - x +c \\ \\ \end{align}

Find   \( \int\ cos^2x\ dx \)

\begin{align} Let \quad I & = \int\ cos^2x\ dx \\ \\ We\ know\ that \quad cos2x & = 2cos^2x-1 \\ \implies cos^2x & = \frac{1}{2}(cos2x+1) \\ \\ Substituting,\ gives \\ \\ I & = \int \frac{1}{2}(cos2x+1)\ dx \\ \\ & = \int\ \frac{1}{2}cos2x\ dx + \int\ \frac{1}{2}\ dx \\ \\ & = \frac{1}{4}sin2x+ \frac{1}{2}x + c\\ \\ \end{align}

Find   \( \int\ sin^2x\ dx \)

\begin{align} Let \quad I & = \int\ sin^2x\ dx \\ \\ We\ know\ that \quad cos2x & = 1- 2sin^2x \\ \implies sin^2x & = \frac{1}{2}(1-cos2x) \\ \\ Substituting,\ gives \\ \\ I & = \int \frac{1}{2}(1-cos2x)\ dx \\ \\ & = \int\ \frac{1}{2}\ dx - \int\ \frac{1}{2}cos2x\ dx \\ \\ & = \frac{1}{2}x - \frac{1}{4}sin2x + c\\ \\ \end{align}

Find   \( \int\ tan^2x\ dx \)

\begin{align} Let \quad I & = \int\ tan^2x\ dx \\ \\ We\ know\ that \quad tan^2x + 1 & = sec^2x \\ \implies tan^2x & = sec^2x - 1 \\ \\ Substituting,\ gives \\ \\ I & = \int (sec^2x-1)\ dx \\ \\ & = \int\ sec^2x\ dx - \int\ 1\ dx \\ \\ & = tanx - x + c\\ \\ \end{align}