# Trig Integration -- secant and tangent problem

• Sep 27th 2009, 02:26 PM
bgonzal8
Trig Integration -- secant and tangent problem
Indefinite integral of (sec(x))^1/2(tan^3(x)).... I know that tan^3(x) becomes tan^2(x) and tanx.... so it will be the Indefinite integral of (sec(x))^-1/2 (tan^2(x)) secx tanx dx ... I dont understand why my book says that there will be a secx in there, why isn't it just (sec(x))^1/2 (tan^2(x)) tanx dx?
• Sep 27th 2009, 03:12 PM
bgonzal8
Quote:

Originally Posted by bgonzal8
Indefinite integral of (sec(x))^1/2(tan^3(x)).... I know that tan^3(x) becomes tan^2(x) and tanx.... so it will be the Indefinite integral of (sec(x))^-1/2 (tan^2(x)) secx tanx dx ... I dont understand why my book says that there will be a secx in there, why isn't it just (sec(x))^1/2 (tan^2(x)) tanx dx?

Nevermind I figured it out... sec-1/2 (x) +sec x = -1/2 +1 = 1/2 .. sec^1/2 (x) is the original factor...
• Sep 27th 2009, 03:26 PM
DeMath
Quote:

Originally Posted by bgonzal8
Indefinite integral of (sec(x))^1/2(tan^3(x)).... I know that tan^3(x) becomes tan^2(x) and tanx.... so it will be the Indefinite integral of (sec(x))^-1/2 (tan^2(x)) secx tanx dx ... I dont understand why my book says that there will be a secx in there, why isn't it just (sec(x))^1/2 (tan^2(x)) tanx dx?

Hint:

$\int {{{\sec }^{1/2}}x\,{{\tan }^3}x\,dx} = \int {\frac{{{{\tan }^3}x}}{{{{\cos }^{1/2}}x}}\,dx} = \int {\frac{{{{\sin }^3}x}}
{{{{\cos }^{1/2}}x\,{{\cos }^3}x}}\,dx} = \int {\frac{{{{\sin }^3}x}}
{{{{\cos }^{7/2}}x}}\,dx} =$

$= - \int {\frac{{\sin x\left( {{{\cos }^2}x - 1} \right)}}
{{{{\cos }^{7/2}}x}}\,dx} = \left\{ \begin{gathered}
\cos x = u, \hfill \\
- \sin x\,dx = du \hfill \\
\end{gathered} \right\} = \int {\frac{{{u^2} - 1}}
{{{u^{7/2}}}}\,du} = \ldots$