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JH
Numerade Educator

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Problem 54 Hard Difficulty

Evaluate the indefinite integral. Illustrate, and check that your answer is reasonable, by graphing both the integrand and its antiderivative (taking $ C = 0 $).

$ \displaystyle \int \sec^4 \left (\frac{1}{2} x \right) dx $

Answer

$$
2 \tan \frac{x}{2}+\frac{2}{3} \tan ^{3} \frac{x}{2}+C
$$

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Video Transcript

here we'd liketo evaluate the integral of sea cans of the fourth power of X over too. And we'LL check the answers reasonable by graphing both the answer grand which is the sequence of the fourth power over here as well as the anti derivative. And then we'LL check that our answer is reasonable by applying the grafts. So first, let's just go ahead and evaluate this thing. One way to proceed is toe rewrite this into brand a Sikh and square times he can square and then at this point, we can apply one of our protagonist identities to write C can't square as tan squared plus one. So let's do that. We have hands where eggs over too plus one and then see counselor So this into grand suggests that we should apply u substitution. Let's go and take you two be tana bets over too. So that do you is C can squared of X over too and by the chain rule left the multiply by one half and then dx equivalently to do you a Sikh and slurred so applying the u substitution weaken right this integral as two times you swear plus one deal which is to you cubed over three plus to you plus e and here we don't week actually ignore the constant of integration by due to the assumption we're allowed to take C to be zero. So let's go ahead and do that. And then the last thing to do is to come back to our use up and replace you with a tangent. So we have to intention cute about silver, too, all over three plus two attention of eggs over too. And that's our answer, Scott. And actually they know this is blue. So they the anti derivatives in blue. The Instagrammed is in red. So let's make one observation here before we go to the graph. So we have. If the anti derivative of the red is equal to the blue, that means that the derivative of the blue so d over DX of two thirds tan cute X over two. So this all implies thiss. So this should all equal the original into Grand seeking to the fourth. So let's check by looking at the grafts if this is true. So this means that the derivative of the blue graph should be the rod graft. So the round graph is giving us the slope of the blue graph. So here we only have the grafts ready. Justus before the Inter Grant is in red as he can forthe and the anti derivative is in blue, so the red graph should be the slope of the tangent line. Since these grafts are basically repeat graphs, let's just focus on here between negative two and two. So here at the origin and the blue graph, it looks like the slope is about one, and that's exactly the value of the red graft. There notice that the blue graph is always increasing, so it means that there's derivative is always positive, and the paragraph here is always positive and notice that as the ex gets closer to two from the left that the graph is rising more more quickly. So that means that the rate of change is increasing and this is what's happening. The red graft, a derivative, is getting larger and larger, and same thing is we're closer to negative, too. Here around negative to the graph is growing really fast, so it means that the derivative should be large and the red graph is indeed large. So these facts suggest that our answer's correct