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Evaluate the integral.

$ \displaystyle \int_0^{\frac{\pi}{4}} \sec^6 \theta \tan^6 \theta d \theta $

$\frac{316}{693}$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 2

Trigonometric Integrals

Integration Techniques

Missouri State University

University of Michigan - Ann Arbor

University of Nottingham

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this problem is from Chapter seven section to problem number twenty six in the book Calculus Early. Transcendental lt's a tradition by shape store. Here we have a definite integral from zero to power for sequence of the six power state of times Tangent to the sixth power of Saito. Since we haven't even power on the sea, can't that's Pollara factor of sea can squared so that we could eventually use the U substitution. So it's rewrite this as c cancer. The fourth power, which we can also write as seek and square squared SATs are seeking to the fourth We have tangents of the six power and then we have another power. She can't square now. Here at the end. Now we can replace C can squared over here in the parentheses with this pig protagonist identity on the right. She can't squared is tan square plus one. Let's use this fact I have an integral zero power before then. In the Prentiss is using the protagonist identity that becomes tan squared data plus one and the whole thing and the parentheses is still being square. Tans of the six and see counselor see there and the reason for doing this was that Now we can use the U substitution take you to be ten Santa So that do you a Sikh and squared data Dita. So if we look at our previous integral we see here we have do you tan to the six will become you to the sixth And over here in parentheses we have useful air plus one and that's also being squared further. Our limits of integration will change because we've done a few substitution. So the lower limit was previously zero. Now that's going to become tanu zero, which is also zero. The upper limit was power before the new upper limit is Tana Power before, which is one. So after the substitution are integral becomes so we have our new limits, Sierra one, Then we have you squared. Actually, at this point, let's actually evaluate this new square plus one square. So let's distribute this tohave. You're the fourth to you square plus one. We have you to the sixth and then do you before we integrate, let's go ahead and distribute issue to the six power through the parentheses. We have you to the tent to you today, plus you to the sixty year and we could use the power of three times here. Its value with these three. So we have you to the lemons over eleven. See you tonight. Nine. Power overnight plus you to the seven over seven, and we have to plug in the end. Points zero and one. So it's plum season. We have won over eleven to over nine one over seven. So this is after we plug in one. And then when we plug in zero, we see that each of these three terms will become zero. So we should subtract zero, and we could simplify and add these three fractions together. Doing so she'LL get a set three sixteen over. Excuse me six ninety three and that's our final answer.

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