Question

Evaluate the Integral\(\int {\frac{{1 - ta{n^2}x}}{{se{c^2}x}}dx} \)\(\)

Short answer

\(\int {\frac{{1 - {{\tan }^2}x}}{{{{\sec }^2}x}}dx} = \frac{1}{2}\sin 2x + c\)

Using the formula\(1 + ta{n^2}x = se{c^2}x\), substitute the value of\(ta{n^2}x\)in the question. Using another identity,\(co{s^2}x = \frac{{1 + cos2x}}{2}\), the integral gets much simpler to solve.

Step-by-step solution

Step 1:Integrating the Equation

\(\begin{aligned}{l}I &= \int {\frac{{1 - ta{n^2}x}}{{se{c^2}x}}dx} \\1 + ta{n^2}x &= se{c^2}x\\ta{n^2}x &= se{c^2}x - 1\\I &= \int {\frac{{1 - \left( {se{c^2}x - 1} \right)}}{{se{c^2}x}}dx} \\I &= \int {\frac{{1 - se{c^2}x + 1}}{{se{c^2}x}}dx} \\I &= \int {\frac{{2 - se{c^2}x}}{{se{c^2}x}}dx} \end{aligned}\)

Divide the integral, both numerator and denominator by \(se{c^2}x\).

\(\begin{aligned}{l}I &= \int {\left( {\frac{2}{{{{\sec }^2}x}} - \frac{{{{\sec }^2}x}}{{{{\sec }^2}x}}} \right)dx} \\I &= \int {\left( {\frac{2}{{{{\sec }^2}x}} - 1} \right)} dx\\I &= \int {\frac{2}{{{{\sec }^2}x}} - \int {1\,} dx} \end{aligned}\)

We know, \(\frac{1}{{se{c^2}x}} = co{s^2}x\).

Therefore, \(I = \int {2co{s^2}x} \,dx - \int {1\,dx} \)

we know, \(co{s^2}x = \frac{{1 + cos2x}}{2}\).

Substitute the value in integral I

\(\begin{aligned}{l}I &= \int {2x\frac{{\left( {1 + cos2x} \right)}}{2}dx} - \int {1\,dx} \\I &= \int {\left( {1 + cos2x} \right)dx - \int {1\,dx} } \\I &= \int {1\,dx + \int {cos2x\,dx - \int {1\,dx} } } \\I &= \int {cos\,2x\,dx} \end{aligned}\)

We know, Integration of \(cos2x = \frac{{sin2x}}{2}\)

Therefore \(I = \frac{1}{2}sin2x + c\)

Hence, \(I = \int {\frac{{1 - ta{n^2}x}}{{se{c^2}x}}\,dx} = \frac{1}{2}sin2x + c\)