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What's the integral of $(\sec x)(\tan x)(1+\sec x)^{1/2}$? How do you solve this by integration by parts?

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3 Answers

Use substitution $u=1+\sec{x}$ then we get $du=\sec{x}\tan{x}\,dx$

$$\int{\sec{x}\tan{x}\sqrt{1+\sec{x}}\,dx}=\int{\sqrt{u}\,du}$$ $$=\frac{2}{3}(u)^{\frac{3}{2}}+C=\frac{2}{3}(1+\sec{x})^{\frac{3}{2}}+C$$

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From the original post: "How do you solve this by integration by parts" ;) –  N. S. Feb 3 '13 at 20:29
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This is an easy substitution problem: $u=1+\sec(x)$.

By parts let $u=\sqrt{1+\sec(x)} dv=sec(x) tan(x)$ and then $du=\frac{\sec(x)\tan(x)}{2 \sqrt{1+\sec(x)}}$ and $v=1+\sec(x)$.


$$\int \sec(x)\tan(x)(1+\sec(x))^{1/2}= (1+\sec(x))^{\frac32}-\frac{1}{2}\int \sec(x)\tan(x)(1+\sec(x))^{1/2}$$

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why don't you just take $u=1+\sec{x}$?i think it would be more simple :) –  A Ricko Maulidar Feb 3 '13 at 19:08
@ARickoMaulidar I mentioned that at the beginning of my solution, but the OP asked explicitely for a solution involving integration by parts ;) –  N. S. Feb 3 '13 at 20:28
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There are generally very few things you can do with roots in integrals. So typically, when you're faced with $\sqrt{\text{something}}$, you make the substitution $u = \text{something}$ so as to remove all of the complexity from underneath root to outside of it, and then try to work from there.

As the other answers have shown, this particular problem becomes extremely easy after doing so.

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