Chapter 6: Techniques of Integration
Section 6.2: Trigonometric Integrals
Example 6.2.10
Evaluate the indefinite integral ∫csc3x ⅆx.
Solution
Mathematical Solution
The given indefinite integral yields to a parts integration, an application of the trig identity cot2x=csc2x−1, and the device of solving for the unknown integral. In particular, take u=cscx so du=−cscxcotx; then dv=csc2x dx so v=−cotx. The complete calculation can be seen in Table 6.2.10(a).
∫csc3x ⅆx
= −cscxcotx−∫cotx⋅cscxcotx dx
= −cscxcotx−∫cot2xcscx dx
= −cscxcotx−∫csc2x−1cscx dx
= −cscxcotx−∫csc3x dx+∫cscx dx
2∫csc3x dx
= −cscxcotx+∫cscx dx
= −cscxcotx+ln(cscx−cotx)
∫csc3x dx
=−12cscxcotx+12ln(cscx−cotx)
Table 6.2.10(a) evaluation of ∫csc3x ⅆx
Maple Solution
Evaluation in Maple
Control-drag the given integral.
Context Panel: Evaluate and Display Inline
∫csc3x ⅆx = −12⁢cos⁡xsin⁡x2+12⁢ln⁡csc⁡x−cot⁡x
Note that Maple expresses cscxcotx in terms of the sine and cosine functions, and that the logarithm is not taken of the absolute value of cscx− cotx. Should this sum be negative, the logarithm will be complex, and a complex constant of integration would compensate.
Table 6.2.10(b) contains a solution generated with the tutor after the Sum, Constant Multiple, and Cosecant rules were selected as Understood Rules.
∫csc⁡x3ⅆx=−csc⁡x⁢cot⁡x−∫csc⁡x⁢cot⁡x2ⅆxparts,csc⁡x,−cot⁡x=−csc⁡x⁢cot⁡x−∫csc⁡x⁢csc⁡x2−1ⅆxrewrite,cot⁡x2=csc⁡x2−1=−csc⁡x⁢cot⁡x−∫csc⁡x3ⅆx−ln⁡csc⁡x+cot⁡xrewrite,csc⁡x⁢csc⁡x2−1=csc⁡x3−csc⁡x=−csc⁡x⁢cot⁡x2−ln⁡csc⁡x+cot⁡x2 [solve
Table 6.2.10(b) Annotated stepwise evaluation of ∫csc3x ⅆx via the Integration Methods tutor
Note that an annotated stepwise solution is available via the Context Panel with the "All Solution Steps" option.
The rules of integration can also be applied via the Context Panel, as per the figure to the right.
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