{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sept04pdf

sept04pdf - Answers to Test 1 given on In 1-7 nd each...

This preview shows pages 1–3. Sign up to view the full content.

Answers to Test 1 given on Sept 15, 2004 In 1-7 find each integral. 1. Z sin 2 3 x dx Use a half angle formula which is sin 2 θ = 1 2 (1 - 2 cos θ ) . So Z sin 2 3 x dx = Z 1 2 (1 - cos 6 x ) dx = 1 2 Z 1 dx - Z cos 6 x dx Look at the second term. We want to use the formula Z cos u du = sin u + C. So u = 6 x du = 6 dx = 1 2 Z 1 dx - 1 6 Z cos 6 x 6 dx = 1 2 x - 1 6 sin 6 x + C = 1 2 x - 1 12 sin 6 x + C 2. Z tan 4 x sec 6 x dx Z tan 4 x sec 6 x dx = Z tan 4 x sec 4 x sec 2 x dx = Z tan 4 x (sec 2 x ) 2 sec 2 x dx = Z tan 4 x (1 + tan 2 x ) 2 sec 2 x dx u = tan x du = sec 2 x dx Substituting 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
= u 4 (1 + u 2 ) 2 du = Z u 4 (1 + 2 u 2 + u 4 ) du = Z ( u 4 + 2 u 6 + u 8 ) du = Z u 4 du + 2 Z u 6 du + Z u 8 du Use formula Z u n du = u n +1 n + 1 = 1 5 u 5 + 2 7 u 7 + 1 9 u 9 + C = 1 5 tan 5 x + 2 7 tan 7 x + 1 9 tan 9 x + C 3. Z x 6 ln x dx . Use integration by parts. u = ln x dv = x 6 dx du = 1 x dx v = Z x 6 dx = x 7 7 uv - Z v du = (ln x ) x 7 7 - Z x 7 7 1 x dx = x 7 7 ln x - 1 7 Z x 6 dx Use formula Z u n du = u n +1 n + 1 = x 7 7 ln x - 1 7 x 7 7 + C Z x 6 ln x dx = x 7 7 ln x - x 7 49 + C 4.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 5

sept04pdf - Answers to Test 1 given on In 1-7 nd each...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online