Calculus Volume 1

Chapter 5 | Integration

603

⌠ ⌡

−1

⎛ ⎝ 2 x 3 +3 x ⎞ ⎠

⎛ ⎝ x 4 +3 x 2 ⎞ ⎠

2 ∫

u −1 du .

dx = 1

Then we have

1 2 ∫

u −1 du = 1

u | + C

2 ln|

= 1 2 ln |

x 4 +3 x 2 | + C .

Example 5.47 Finding an Antiderivative of a Logarithmic Function

Find the antiderivative of the log function log 2 x .

Solution Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have ∫ log 2 xdx = x ln2 (ln x −1)+ C .

Find the antiderivative of log 3 x .

5.39

Example 5.48 is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration. Example 5.48 Evaluating a Definite Integral

Find the definite integral of ⌠ ⌡ 0 π /2

sin x 1+cos x

dx .

Solution We need substitution to evaluate this problem. Let u =1+cos x , , so du =−sin xdx . Rewrite the integral in terms of u , changing the limits of integration as well. Thus, u =1+cos(0)=2 u =1+cos ⎛ ⎝ π 2 ⎞ ⎠ =1.

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