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2 years ago

12

Please help!!!

1 answer:

marusya05 [52]2 years ago

5 0

Answer:

the answer is B.FALSE

Step-by-step explanation:

hope it helps

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$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$

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$\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c$

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2 years ago

I need to solve: 6x+3^59

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3 years ago