# Calculating $\int_{0}^{\sqrt{3}} \frac{dt}{\sqrt{1+t^2}}$.

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?

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

Second, the correct antiderivative here is $$\sinh^{-1}t$$ and so the answer is $$\sinh^{-1}\sqrt3$$.
• Thanks. I used $t=\sinh{(x)}$. A quick question why is $\sqrt{\sec^2{(x)}}=|\sec{(x)}|\neq \sec{(x)}$ but $\sqrt{\cosh^2{(x)}}=\cosh{(x)}$. – squenshl Aug 14 at 8:15
• More simply, it is $\ln(\sqrt 3+2)$. – Bernard Aug 14 at 9:09
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 $$u=\tan x+\sec x$$.
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 $$\sec(x)+\tan(x)$$.