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Evaluate the integral.

$ \displaystyle \int_0^{\frac{\pi}{2}} \frac{\cos t}{\sqrt{1 + \sin^2 t}}\ dt $

$\int_{0}^{\pi / 2} \frac{\cos t}{\sqrt{1+\sin ^{2} t}} d t=\ln (\sqrt{2}+1)$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

Integration Techniques

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here we have a definite integral from zero to pie over two of co sign of tea divided by the square root of one plus sign square. Let's do it. Use up here and let's take you to be sci fi Then do you is cosign TT Since this is a definite no girl, we should switch these limits of integration So the new lower limit will become using this formula. Pierre. So we plug in t equals zero and we have u equals sign zero, which is zero and then for the upper limit. Piper too. So we have upper limit you equals signed by over two, which is one and then by Are you sub cosign t dt is just do you and then the denominator we have one plus you square inside the radical. So now for this new integral we should do entrance up. So let's take you equals Tan Daito Then do you a sequence where data and once again because we have Ah, definitely, girl. We should go out and switch these limits. So originally we had zero for you also Wait. One more observation since this is a definite no girl. Oh, when you're used when your tricks up is of the form eight and data the requirement is that date is between negative five for two and powerful too. So plugging in you equal zero to get a new lower limit, we have zero equals tan data. And if they does in this interval appear the only time tangent could be zero is a data equals zero. So that's our new lower limit for the upper limit. We have one equals tan data and the only time this happens in our honorable negative Piper tutto perverts Who is that power for some running on room here. So let me go to the next page. So we had that was the general. So now, after using our tricks up, we have sealed up IRA for do you with sequins. Where? Data. And in the bottom we have tan squared plus one. So using the Pythagorean identity for Tangent Square, we have seeking squared inside the radical and then the square root of seek and square as you see again so we could cross off one of the sequence Sierra Pi over four seek and data and we know this integral to be natural log, absolute value seek and data plus tan data. So this is a trick in a girl and there end points or zero power before. So if we plug in pi over four first, seeking a pi over four is rude, too. Tangent on power for is one and then seek in of zero is one. Tension of zero is zero and then natural log of one zero so we could ignore the second term and their final answer. Natural log one plus radical too. And you could lose the absolute value because one plus radical to his positive number.

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