Intergration With Trigonometry

**twoteenine** · March 16th 2010, 06:30 PM

$\int \frac{\sec^5 \theta}{\tan^2\theta}d\theta$

I'm having troubles starting this. I've tried substituting for tangent,splitting up the secants, and a couple other methods but I get stuck after a couple steps. How should i start out?

**Prove It** · March 16th 2010, 07:07 PM

Originally Posted by twoteenine

$\int \frac{\sec^5 \theta}{\tan^2\theta}d\theta$

I'm having troubles starting this. I've tried substituting for tangent,splitting up the secants, and a couple other methods but I get stuck after a couple steps. How should i start out?

integral[((Sec[x])&# x5e;5)/((Tan[x]&#x29 ;^2)] - Wolfram|Alpha

Click Show Steps.

**simplependulum** · March 17th 2010, 02:07 AM

$\frac{ \sec^5(x) }{ \tan^2(x) }$

$= \frac{ \sec^4(x) }{ \tan^2(x) } \cdot \sec(x)$

$= \frac{ ( \tan^2(x) + 1)^2 }{ \tan^2(x) }\cdot \sec(x)$

$= \frac{ \tan^4(x) + 2\tan^2(x) + 1}{\tan^2(x) } \cdot \sec(x)$

$= [ \tan^2(x) + 2 + \frac{1}{\tan^2(x)} ] \cdot \sec(x)$

$= \tan^2(x)\sec(x) + 2\sec(x) + \frac{1}{ \tan(x) \cdot \tan(x) \cdot \cos(x) }$

$= \tan^2(x)\sec(x) + 2\sec(x) + \csc(x)\cot(x)$

its integral $= \int \tan^2(x)\sec(x) ~dx + 2\int \sec(x)~dx + \int \csc(x)\cot(x) ~dx$

$= \int \tan^2(x)\sec(x) ~dx + 2\int \sec(x)~dx - \csc(x) + C$

Consider $I = \int \tan^2(x)\sec(x) ~dx$

$I = \int [ \sec^3(x) - \sec(x) ]~dx$ $(1)$

but integration by parts gives

$I = \sec(x) \cdot \tan(x) - \int \sec^3 (x)~dx$

therefore ,

$\int \sec^3(x) ~dx = \sec(x) \tan(x) - I$ $(2)$

By substituting $(2)$ into $(1)$

we have

$I = \sec(x) \tan(x) - I - \int \sec(x)~dx ~~~ \implies$

$I = \frac{1}{2} \sec(x) \tan(x) - \frac{1}{2} \int \sec(x)~dx$

The integral

$\int \frac{ \sec^5(x) }{ \tan^2(x) } ~dx =$

$\frac{1}{2} \sec(x) \tan(x) + \frac{3}{2} \int \sec(x)~dx - \csc(x) + C$

$= \frac{1}{2} \sec(x) \tan(x) + \frac{3}{2} \cosh^{-1}[\sec(x)] - \csc(x) + C$

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