{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

q6-sol-1042-sp08

q6-sol-1042-sp08 - Calculus II-1042 Solution of Quiz 6 Fri...

This preview shows page 1. Sign up to view the full content.

Calculus II—1042 Solution of Quiz # 6 Fri., 03/07/08 1. Z π 0 8 cos 4 (2 x ) dx . Solution: Since cosine has EVEN exponent so break the even power using cos 2 x = 1 2 + cos(2 x ) 2 . We continue to use it till no even power left. First we do some algebra to break the even powers of cos 4 (2 x ) then we compute its anti-derivative 8 cos 4 (2 x ) = 8 ( cos 2 (2 x ) ) 2 = 8 1 2 + cos(4 x ) 2 2 = 8 4 ( 1 + 2 cos(4 x ) + cos 2 (4 x ) ) = 2 1 + 2 cos(4 x ) + 1 2 + cos(8 x ) 2 = 2 3 2 + 2 cos(4 x ) + cos(8 x ) 2 = 3 + 4 cos(4 x ) + cos(8 x ) Z π/ 8 0 8 cos 4 (2 x ) dx = Z π/ 8 0 (3 + 4 cos(4 x ) + cos(8 x )) dx = 3 x + 4 4 sin(4 x ) + 1 8 sin(8 x ) | π/ 8 0 = 3 π 8 + sin( π 2 ) + 1 8 sin( π ) - 0 + sin(0) + 1 8 sin(0) = 3 π 8 + 1 2. Z sin 4 y cos 3 y dy . Solution: Since cos y has ODD power so split one cos y and convert the remaining cos 2 y into sin y using cos 2 y = 1 - sin 2 y , then make a substitution u = sin y Z sin 4 y cos 3 y dy = Z sin 4 y · cos 2 y · cos y dy = Z sin 4 y · (1 - sin 2 y ) · cos y dy Substutution: u = sin y, du = cos y dy = Z u 4 (1 - u 2 ) du = Z ( u 4 - u 6 ) du = u 5 5 - u 7 7 + C = sin 5 y 5 - sin 7 y 7 + C 3. Z sec 4 t tan 5 t dt . Solution: Observe that tan t has ODD exponent and sec t has EVEN exponent. We can start with anyone of them. Sine the exponent of sec t is smaller than exponent of tanx so working with sec t will be easier in this problem. Method 1: Considering EVEN exponent of sec
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern