The Notes: `int tan(x) sec^3(x) dx` Evaluate the integral

Friday, September 11, 2009

$\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$

To solve, apply u-substitution method.

Let the u be:

$\displaystyle{u}={\sec{{\left({x}\right)}}}$

Then, differentiate it.

$\displaystyle{d}{u}={\tan{{\left({x}\right)}}}{\sec{{\left({x}\right)}}}{\left.{d}{x}\right.}$

To be able to plug-in this to the integral, re-write the integrand as:

$\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}={{\sec}}^{{2}}{\left({x}\right)}\cdot{\tan{{\left({x}\right)}}}{\sec{{\left({x}\right)}}}{\left.{d}{x}\right.}$

Then, express the integrand in terms of u. So it becomes:

$\displaystyle=\int{{u}}^{{2}}{d}{u}$

To take the integral of this, apply the formula $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle=\frac{{{u}}^{{3}}}{{3}}+{C}$

And, substitute back u= sec(x).

$\displaystyle=\frac{{{{\sec}}^{{3}}{\left({x}\right)}}}{{3}}+{C}$

Therefore, $\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}=\frac{{{{\sec}}^{{3}}{\left({x}\right)}}}{{3}}+{C}$ .

The Notes

Friday, September 11, 2009

$\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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