Rational functions of trigonometric functions | Ch 7.4 - 76E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

ISBN: 9780321570567 2

Solution for problem 76E Chapter 7.4

Calculus: Early Transcendentals | 1st Edition

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Problem 76E

Rational functions of trigonometric functions

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution u =tan (x/2) or x = 2 tan−1 u. The following relations are used in making this change of variables.

Evaluate .

Step-by-Step Solution:

Problem 76E

Rational functions of trigonometric functions
An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution $u\ =tan\ (x/2)$ or $x\ =\ 2\ arctan\ u.$
The following relations are used in making this change of variables:
A: $dx=\ \frac{2}{1+{u}^{2}}du$
B: $sinx=\ \frac{2u}{1+{u}^{2}}$
C: $cosx=\ \frac{1-{u}^{2}}{1+{u}^{2}}$

Evaluate $\int\limits_{\ }^{\ }\frac{dx}{1+sinx+cosx}$
Solution:-
Step 1
$I=\int\limits_{\ }^{\ }\frac{dx}{1+sinx+cosx}$ (Making the indicated substitutions in the integral)
= $\int\limits_{\ }^{\ }\frac{\frac{2}{1+{u}^{2}}}{1+\frac{2u}{1+{u}^{2}}+\frac{1-{u}^{2}}{1+{u}^{2}}}du$
= $\int\limits_{\ }^{\ }\frac{\frac{2}{1+{u}^{2}}}{\frac{1+{u}^{2}+2u+1-{u}^{2}}{1+{u}^{2}}}du$
= $\int\limits_{\ }^{\ }\frac{2}{2+2u}du$
= $\int\limits_{\ }^{\ }\frac{1}{1+u}du$