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Calculate $\int_{0}^{\sqrt{3}} \frac{dt}{\sqrt{1+t^2}}$.
So we let $t = \tan{(x)}$ so $1+t^2 = 1+\tan^2{(x)} = \sec^2{(x)}$ which means $$\int \frac{dt}{\sqrt{1+t^2}} = \int \frac{1}{\sec^2{(x)}}\times \sec^2{(x)} \; dx = x+C = \tan^{-1}{(t)}+C.$$
Is this good?

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    $\begingroup$ No, you're given a definite integral. Final result must be a number.. $\endgroup$ – ganeshie8 Aug 14 at 7:32
  • $\begingroup$ Correct. I meant the indefinite integral. $\endgroup$ – squenshl Aug 14 at 7:33
  • $\begingroup$ Hey wait, in the denominator how did you get $\sec^2x$ ? $\endgroup$ – ganeshie8 Aug 14 at 7:33
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    $\begingroup$ $t=\sinh{x}$ helps. $\endgroup$ – Michael Rozenberg Aug 14 at 7:34
  • $\begingroup$ Agreed I didn't take the square root on the bottom oops! $\endgroup$ – squenshl Aug 14 at 7:36
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First, as mentioned in comments, since this is a definite integral there should not be an integration constant at the end.

Second, the correct antiderivative here is $\sinh^{-1}t$ and so the answer is $\sinh^{-1}\sqrt3$.

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  • $\begingroup$ Thanks. I used $t=\sinh{(x)}$. A quick question why is $\sqrt{\sec^2{(x)}}=|\sec{(x)}|\neq \sec{(x)}$ but $\sqrt{\cosh^2{(x)}}=\cosh{(x)}$. $\endgroup$ – squenshl Aug 14 at 8:15
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    $\begingroup$ More simply, it is $\ln(\sqrt 3+2)$. $\endgroup$ – Bernard Aug 14 at 9:09
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let $t = \tan{(x)}$ so $1+t^2 = 1+\tan^2{(x)} = \sec^2{(x)}$ which means $$\int_0^{\sqrt3} \frac{dt}{\sqrt{1+t^2}} = \int_0^{\frac\pi3} \frac{1}{\sec{(x)}}\times \sec^2{(x)} \; dx=\int_0^{\frac\pi3}\sec(x)\,dx=\log(\tan x+\sec x)\mid_0^{\frac\pi3}.$$ Here we used ($x\in[0,\frac\pi2]$) $$\int\sec x\,dx=\int\frac{\sec^2x+\tan x\sec x}{\tan x+\sec x}\,dx=\int\frac1u\,du=\log u+C=\log(\tan x+\sec x)+C$$ where $u=\tan x+\sec x$.

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Your result is not correct, because $\sqrt{\sec^2(x)}=|\sec(x)|\neq \sec^2(x)$. And to integrate $\sec(x)$, multiply the numerator and denominator by $\sec(x)+\tan(x)$.

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