121616949-math.175

121616949-math.175 - x are easy The cos 3 2 x integral is...

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

8.2 Powers of sine and cosine 161 Now use u = cos x , du = - sin x dx : Z sin x (1 - cos 2 x ) 2 dx = Z - (1 - u 2 ) 2 du = Z - (1 - 2 u 2 + u 4 ) du = - u + 2 3 u 3 - 1 5 u 5 + C = - cos x + 2 3 cos 3 x - 1 5 cos 5 x + C. EXAMPLE 8.6 Evaluate Z sin 6 x dx . Use sin 2 x = (1 - cos(2 x )) / 2 to rewrite the function: Z sin 6 x dx = Z (sin 2 x ) 3 dx = Z (1 - cos 2 x ) 3 8 dx = 1 8 Z 1 - 3 cos 2 x + 3 cos 2 2 x - cos 3 2 x dx. Now we have four integrals to evaluate: Z 1 dx = x and Z - 3 cos 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x are easy. The cos 3 2 x integral is like the previous example: Z-cos 3 2 x dx = Z-cos 2 x cos 2 2 x dx = Z-cos 2 x (1-sin 2 2 x ) dx = Z-1 2 (1-u 2 ) du =-1 2 ± u-u 3 3 ¶ =-1 2 ± sin 2 x-sin 3 2 x 3 ¶ ....
View Full 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