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JH
Numerade Educator

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Problem 31 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int \tan^5 x dx $

Answer

$\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2} -\ln |\cos x|+C$

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Video Transcript

This problem is from Chapter seven, a section to problem number thirty one from the book Calculus Early Transcendental Sze eighth Edition by James Door and we have Indefinite and a roll of tangents of Fifth Power Vex. One way to proceed here is tow Brake This tangent to the Fifth Power by writing It is tangent, Cubed Time, Stanton Square and for the tangent squared. We can use the patellar, even identity, over here on the right to rewrite this as seek and squared X minus one. So doing so gives us ten Q picks, time seeking swear legs minus can cubed times one. It looks like for this first integral here, we can use a U substitution because we have a C can square, not quite for the second and the girl, at least not yet. So let's go ahead and break this into two separate individuals. So here, let's get color coordinated. So let's do the first integral in Blue Team Cube seek and square minus, and we have integral tension Cube. And let's go ahead and write That tangent. Cubed is Tan Square Time's tangent, and the reason for doing so is once again we can rewrite this tan squared as seeking squared minus one for first Integral in blue, we see that we can use a new substitution. Let's take you to be a tangent, then do you? A Sikh and squares GX. So let's make pick up our equality down here. So after this, you substitution, we have you cubed, do you? And we still have the second integral here, which, after using this protagonist identity, we have seeking Squared X minus one time's changing for this first integral. We can go ahead and just use the power rule, and we get used to the fourth power over four for the second and the girl. Let's go out and distribute attention through the parentheses in Let's split This up into two and a girl So we have a minus in a rule. Tangin X time seeking square X. So watch out here for the two minuses I wonder with a plus in a girl off tangent times one which is just tangent. So we'LL have to evaluate the noble of tangent. You might have memorized this one, but if not you Khun evaluated by just writing tangent assigned over co sign and then doing the use of for this one. So for this inaugural here, you could go ahead and try you two be co scientifics so that negative do you is sine x dx. So coming back to our inner rule, we could back substitute. But for now, we have more work to do. Let's go ahead and just leave this in terms of you from now. If you want you to the fourth over for minus integral here, we can rely on the same U substitution that we use earlier. Because we have the Sikh and square. So, Scott, and rely on the same you somebody's before. So this becomes you. Deal. And for this final integral. Here we have a minus. Do you see this minus over here? After you so and a girl One over co sign, which is you to you. And we could evaluate this first interval here using the power rule for this one. We could also use the power rule. But this is you to the minus one. So they're the inner world is going to be the natural log of the absolute value of you. So we have you forthe over for minus. You squared over two mine This natural log, absolute value you and plus our constancy of integration. So this step we've evaluated the inaugural. Now it's timeto get our answer back in terms of the variable X. So this is where we were lying. Are you substitution Sze? Let's just be cautious here. The first to use corresponded to our original u sub. Where is the final? You corresponded to the other use of that We used co sign so something to watch out for their So we have tangents of the four power of X over four minus tangent, square of X over two, minus natural log absolute value This you and red That's the coastline and plus our constancy and there's our answer