121616949-math.176 - 9 Z sec 2 x csc 2 x dx ⇒ 10 Z tan 3...

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162 Chapter 8 Techniques of Integration And finally we use another trigonometric identity, cos 2 x = (1 + cos(2 x )) / 2: Z 3 cos 2 2 x dx = 3 Z 1 + cos 4 x 2 dx = 3 2 x + sin 4 x 4 . So at long last we get Z sin 6 x dx = x 8 - 3 16 sin 2 x - 1 16 sin 2 x - sin 3 2 x 3 + 3 16 x + sin 4 x 4 + C. EXAMPLE 8.7 Evaluate Z sin 2 x cos 2 x dx . Use the double angle formulas sin 2 x = (1 - cos(2 x )) / 2 and cos 2 x = (1 + cos(2 x )) / 2 to get: Z sin 2 x cos 2 x dx = 1 4 Z 1 - cos 2 (2 x ) dx = 1 4 Z sin 2 (2 x ) dx = 1 8 Z 1 - cos(4 x ) dx = 1 8 x - sin(4 x ) 4 + C. Exercises Find the antiderivatives. 1. Z sin 2 x dx 2. Z sin 3 x dx 3. Z sin 4 x dx 4. Z cos 2 x sin 3 x dx 5. Z cos 3 x dx 6. Z sin 2 x cos 2 x dx 7. Z cos 3 x sin 2 x dx 8. Z
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Unformatted text preview: 9. Z sec 2 x csc 2 x dx ⇒ 10. Z tan 3 x sec x dx ⇒ 8.3 Trigonometric Substitutions So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. Occasionally it can help to replace the original variable by something more complicated. This seems like a “reverse” substitution, but it is really no different in principle than ordinary substitution....
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