Calculus Volume 1

Answer Key

843

−1 ⎛

⎞ ⎠ + C . Taking C =0 recovers the definite integral over ⎡ ⎣ 2, 6 ⎤ ⎦ .

⎝ x 2

The antiderivative is 1

2 sec

419 .

The general antiderivative is tan −1 ( x sin x )+ C . Taking C =−tan −1 (6sin(6)) recovers the definite integral. 421 .

The general antiderivative is tan −1 (ln x )+ C . Taking C = π

−1 ∞

recovers the definite integral.

2 = tan

423 . sin −1 ⎛ ⎝ e ⎠ + C 425 . sin −1 (ln t )+ C 427 . − 1 2 ⎛ ⎝ cos −1 (2 t ) t ⎞ ⎞ ⎠

2

+ C

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