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Numerade Educator



Problem 12 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_0^2 \frac{dt}{\sqrt{4 + t^2}} $


$\int_{0}^{2} \frac{d t}{\sqrt{4+t^{2}}}=\ln (1+\sqrt{2})$


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Video Transcript

let's compute the integral from zero to of one over the square root of four plus T square. Since the expression in the radical this four plus T squared is of the form a square plus B square, we should use a trick so T equals eight and data in our problem. Since a square is for of the four over here, that means a is too, so that our tricks up should be t equals to tan data taking the differential on both sides. DT equals two sinkings were data potato. Also for this problem, this is a definite integral so we have limits. So if we want, we could find the limits of integration for the new integral in terms of data. So the lower limit which corresponds to t equals zero. We want to find out what data is. So we'LL use this equation up here, so plug in T equals zero. Also, let's make observation when you use a trick substitution on this form, we have the condition that fate is between negative pie over too and pie over too. So this is part of the trick substitution for attention. So we haven't from this equation we see that tangent zero. And this means that data equals zero because the only time tangent zero from negative high over to toe Palmer to is when data equals zero. So for our new integral, the lower limit will still be zero not for the upper limit, which was originally too. We take Teeples too and we plug it into this equation up here. So we have two equals to attention, which means that tangent equals one. And this implies that data is pi over four. Because the only time the tangent function is one in this interval appear is that pie of reform. So that's our new upper limit. Yes, but that then on the numerator, we have GT for marching sub. We know what this is two times c can't square and let's go to the side and simplified this denominator. We have square root of four plus t squared. This's spur of four plus four changes squared and running out of room here has come down here. Let's pull up the four outside the radical that becomes a too. We have one plus tangent squared data. We know this is two times seek and squared data in the square root of that, and the square root of seeking square is simply seeking. So this is our denominator. We could go ahead and cancel off these twos and one of the sea cans. So let's go to the next page of running out of room here, picking up where we left off after cancellation. We have the integral c can't data, but and we know this is the natural log of absolute value of C Can't plus tangent and now ten points zero power for So let's go ahead and plug those in seeking a partner before square of two. Detention apart before is one. And then when we plug it, zero sikandar zero is one and tangent zero zero No, and we know the natural log of born a zero so we could ignore this term. And since this expression over here in parentheses is positive, we could even drop the absolute values so we can write the answer as natural log one plus square, too. And that's our answer