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7-157.

Integrate. Show your steps—if you use u-substitution be sure to change the bounds of integration.

$\int _ { \pi / 4 } ^ { 0 } \sqrt { \operatorname { sin } x } \cdot \operatorname { cos } x d x$
Step 1(a):
$U =\operatorname{sin}x$
$dU =\operatorname{cos}x\ dx$
Step 2(a):
$U\left ( \frac{\pi }{4} \right )=$ lower bound
$U(0) =$ upper bound
Step 3(a):
Assemble the integral and then solve.

$\int \frac { 1 } { \operatorname { sin } x } \operatorname { cos } x d x$
Hint (b):
$U =\operatorname{sin}x$
$dU =\operatorname{cos}x\ dx$

$\int \frac { 3 x ^ { 4 } - 4 x ^ { 2 } - 11 x + 6 } { x ^ { 2 } } d x$
Hint (c):
Before you integrate, rewrite the integrand.
$= \int 3x^{2}-4-\frac{11}{x}+\frac{6}{x^{2}}dx=$

$\int \frac { 1 } { | x | \sqrt { x ^ { 2 } - 1 } } d x$
Answer (d):
$\operatorname{sec}^{−1}x + C$ (this is a special case)

$\int _ { 1 } ^ { e } \frac { 1 } { x } \operatorname { cos } ( \operatorname { ln } x ) d x$
Hint (e):
$\frac{d}{dx}\left ( \frac{1}{x} \right )=\text{ln}x$

$\int _ { - 1 } ^ { 3 } x ^ { 2 } ( x ^ { 3 } - 8 ) d x$
Hint (f):
You could use U-Substitution.
Or you could expand the integrand and evaluate.

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