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How to integrate $$∫\frac{\sec^2x}{(\sec x+\tan x)^{9/2}} dx$$

In my book it is done like

$$\begin{align} \sec x+\tan x &=t \\ \sec x−\tan x &= \frac 1t \\ (\sec x \tan x+\sec^2x)dx &=dt \\ \sec x(\sec x+\tan x)dx &=dt \\ \sec xdx &=\frac 1t dt \\ t+\frac 1t &=2\sec x \end{align}$$

and then it is substituted in the original expression and integrated.

The substitutions seem pretty unintuitive to me. Please share the approach you would follow :-)!

Hosein Rahnama
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    The expression they substituted for appears in the denominator, so it's not that unintuitive. – GGG Jan 04 '16 at 13:00
  • What makes you think that it is unintuitive? – Aditya Agarwal Jan 04 '16 at 13:02
  • letting the denominator to be $t$ is one of typical skills in integration. – dust05 Jan 04 '16 at 13:04
  • Alright.It's not so unintuitive.I agree @AdityaAgarwal...but its just that I wanted to learn some alternative methods ;-)!Because the people in Math SE always surprise me with some new integration tricks :-)! –  Jan 04 '16 at 13:06
  • @user276387 well yes now it does seem okay :-)!I've been out of touch with integration for quite some time..maybe that's why.Will i delete it? –  Jan 04 '16 at 13:07
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    I don't think that you should delete it. – Aditya Agarwal Jan 04 '16 at 13:07

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