Question

To prove the reduction formula, use integration by parts. \(\int {{{\tan }^n}} xdx = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}} xdx\)

Short answer

We proved that\(\)\(\)\(\int {{{\tan }^n}} xdx = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}} xdx\)

Step-by-step solution

Given Equation

To prove that \(\int {{{\tan }^4}} xdx = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}} dx\) \((n \ne 1)\)

Evaluating L.H.S.

Evaluating integration:

Let \({I_n} = \int {{{\tan }^n}} xdx\)

Multiplying & dividing by \({\tan ^2}x\)

=\(\int {\frac{{{{\tan }^n}x \times {{\tan }^2}x}}{{{{\tan }^2}x}}} dx\)

=\(\int {{{\tan }^n}} x \times {\tan ^2}x \times {\tan ^{ - 2}}xdx\)

=\(\int {{{\tan }^{n - 2}}} x{\tan ^2}xdx\) )

=\(\int {{{\tan }^{n - 2}}} x(\sec x - 1)dx\)

\( \Rightarrow {I_n} = \int {{{\tan }^{n - 2}}} x{\sec ^2}xdx - \int {{{\tan }^{n - 2}}} xdx\)

Evaluating Further

Evaluating \(\int {{{\tan }^{n - 2}}} x{\sec ^2}xdx\)

Let \(\tan x = t\)

Differentiating with respect to ‘x’

\({\sec ^2}x = \frac{{dt}}{{dx}}\)

\(\therefore \int {{{\tan }^{n - 2}}} x{\sec ^2}xdx = \int {{t^{n - 2}}} dt\)

=\( = \frac{{{t^{n - 2 + 1}}}}{{n - 2 + 1}}\)

\( \Rightarrow \int {{{\tan }^{n - 2}}} x{\sec ^2}xdx = \frac{{{t^{n - 1}}}}{{n - 1}}\)

Replace t by \(\tan x\)

Substitute this value in Equation (1)

Replace \({I_n}\) by \(\int {{{\tan }^n}} xdx\) and \({I_{n - 2}}\) by\(\int {{{\tan }^{n - 2}}} xdx\)

Hence proved

The answer is \(\int {{{\tan }^n}} xdx = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}} xdx\)