1. integration of x.secx.tanx

how would you integrate x.secx.tanx dx??
thanks for any help

2. $\displaystyle \int x\sec{x}\tan{x}~dx = \int x(\sec{x})'~dx = x\sec{x} - \int \sec{x}~dx$

It can be easily shown that $\displaystyle \int \sec{x}~dx = \ln|\sec{x}+\tan{x}|$ by multiplying the integral expression with $\displaystyle \frac{\sec{x}+\tan{x}}{\sec{x}+\tan{x}}$

3. tanx

what happened to the tanx???

4. Recall that:

$\displaystyle \frac{d}{dx} \sec{x}= (\sec{x})' = \sec{x}\tan{x}$

I wrote $\displaystyle \sec{x}\tan{x}$ in an equivalent form $\displaystyle (\sec{x})'$ as it is easier and faster to see what u and v when using integration by parts.

5. $\displaystyle \int x\cdot sec(x)tan(x)dx$

Use parts, Let $\displaystyle u=x, \;\ dv=sec(x)tan(x)dx, \;\ du=dx, \;\ v=sec(x)$

We get:

$\displaystyle xsec(x)-\int sec(x)dx$

Now, it's easier, huh?.

6. thanks

thanks to both of you i was having trouble seeing that one

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xsecx

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