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$[\mathbf{M}]$ (Calculus required) $^{3}$ Recall from calculus that integrals such as$\int\left(5 \cos ^{3} t-6 \cos ^{4} t+5 \cos ^{5} t-12 \cos ^{6} t\right) d t$are tedious to compute. (The usual method is to apply inte- gration by parts repeatedly and use the half-angle formula.) Use the matrix $P$ or $P^{-1}$ from Exercise 17 to transform $(10) ;$ then compute the integral.

$-6 t+\frac{55}{8} \sin t-\frac{69}{16} \sin 2 t+\frac{15}{16} \sin 3 t-\frac{3}{4} \sin 4 t+\frac{1}{16} \sin 5 t-\frac{1}{16} \sin 6 t+C$

Calculus 3

Chapter 4

Vector Spaces

Section 7

Change of Basis

Vectors

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uh, here we can define new auction to UT equal to fine minus for co sign T s so that we have you to the one over for your aunt. A grand Do you is, um, Minus four times minus sign of P d d or just ah, poor scientist e greedy. And so we conclude the time t d t. Is. Do you buy four? So we end up integrating with respect to you. 5/4 you to the one over four using our formula for the, uh, our formula Franci derivatives. You have that The anti derivative of I bow before you to the one over four is just you to the part where I work. And now for the bounds. For evaluating the time you read it. We have that zero. Gonzo is five lines for transco. Sand zeroes. Five minutes. Four times. One is one. So we subtract one. Uh, well, we value we. It's untracked B and derivative. Evaluated at one with the anti you derivative Evaluated G a pie. Just five minus four. Co sign chi and she's five minus four times minus one riches. But this word is nine. That's your final answer. then is not into the five over four minus one of the five over four. Since the square root of nine is three, another five over four is three to the five over, too. And, of course, from this recent charge, one of the Bible for just long we see that this answer created five over to mine. It's on.

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