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Evaluate the integral.

$ \displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \csc^4 \theta \cot^4 \theta d \theta $

$$\frac{12}{35}$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 2

Trigonometric Integrals

Integration Techniques

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this problem is from Chapter seven, Section two, Problem number thirty eight and the book Calculus. Early Chance in Denzel's A Tradition by James Door. So for this in a roll, we haven't even power on the CO C can't. So let's proceed by pulling out a power of seek and square from the Inter Grant. And then we can use our protagonist identity to rewrite this first Kosi can square. We don't need to do that for the second one because we'LL see that this term will pop out once we do a U substitution and a girlfriend pyre afford a pie over too. So relying on our pathetic unidentified over here, weaken right, this is Cho Tangent squared, plus one time's contention to the Force Time's Corsicans work. Now we see we can apply u substitution. Let's take you to be co attention so that negative do you? No is cause he can't square theater Dita Also, because we have a definite in general, our limits of integration will change. So our new lower limit of integration will become call attention of power for which is one in our upper limit will become Call attention apart over too which is zero. So after applying our U substitution, we have negative. This is coming from the negative and the U substitution integral from one zero you squared plus one. And from this term, we have used to the fourth. And that's our deal right there. So at this point, we can go ahead and using the properties of immigration. We can give her this negative sign if we switched the limits of integration. So let's right, this is the positive. Integral from zero one. And let's go ahead and distribute this you to the fourth through the parentheses. Well, and now we can go ahead and use the power rule to evaluate this. Okay, Plugging in one first for you. And then if we plug in zero, we see that you two the seventh and you to the fifth or both zero. So we just subtract zero and we can go ahead and combined those fractions and we should get twelve over thirty five. And there's our answer

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