Chapter 6: Techniques of Integration
Section 6.2: Trigonometric Integrals
Example 6.2.2
Evaluate the indefinite integral ∫cos4x ⅆx.
Solution
Mathematical Solution
An evaluation of the given integral can be based on the trig identity cos2x=1+cos2 x/2, which must be invoked twice during the course of the computation in Table 6.2.2(a).
∫cos4x ⅆx
= ∫1+cos2 x22 ⅆx
= 14∫1+2 cos2 x+cos22 x dx
= 14∫1 dx+14∫2 cos2 x dx+14∫cos22 x dx
= x4+14 sin2 x+14∫1+cos2⋅2 x2 dx
= x4+14 sin2 x+18∫1 dx+18∫cos4 x dx
= x4+14 sin2 x+x8+132 sin4 x
= 38 x+14 sin2 x+132 sin4 x
Table 6.2.2(a) Solution based on the identity cos2x=1+cos2 x/2
Alternatively, the first reduction formula in Table 6.2.3, namely,
∫cosmx ⅆx=1mcosm−1xsinx+m−1m ∫cosm−2x ⅆx
can be used with m=4 and then again with m=2, as shown in the following computation,
=14 cos3xsinx+34∫cos2x dx
=14 cos3xsinx+3412cosxsinx+12∫1 dx
=14 cos3xsinx+38 cosxsinx+38 x
It would appear that there is less chance of making an arithmetic error in the second solution, the one based on the reduction formula. However, the two solutions are the same: the first solution can be transformed to the second by repeated application of the trig identity sin2 x=2 sinxcosx.
Maple Solution
Evaluate the given integral
Control-drag the integral.
Context Panel: Evaluate and Display Inline
∫cos4x ⅆx = 14⁢cos⁡x3⁢sin⁡x+38⁢cos⁡x⁢sin⁡x+38⁢x
Table 6.2.2(b) contains an annotated stepwise solution generated with the tutor. The steps in Table 6.2.2(b) are a combination of Maple suggestions and human intervention. In particular, both the Sum and Constant Multiple rules were checked as Understood Rules to shorten the computation.
∫cos⁡x4ⅆx=∫1+cos⁡2⁢x2ⅆx4rewrite,cos⁡x2=12+cos⁡2⁢x2=∫cos⁡2⁢x2ⅆx4+∫cos⁡2⁢xⅆx2+∫1ⅆx4rewrite,1+cos⁡2⁢x2=cos⁡2⁢x2+2⁢cos⁡2⁢x+1=∫cos⁡2⁢x2ⅆx4+∫cos⁡2⁢xⅆx2+x4constant=∫cos⁡u2ⅆu8+∫cos⁡2⁢xⅆx2+x4change,u=2⁢x,u=∫cos⁡2⁢uⅆu16+∫12ⅆu8+∫cos⁡2⁢xⅆx2+x4rewrite,cos⁡u2=cos⁡2⁢u2+12=∫cos⁡2⁢uⅆu16+u16+∫cos⁡2⁢xⅆx2+x4constant=∫cos⁡2⁢uⅆu16+3⁢x8+∫cos⁡2⁢xⅆx2revert=∫cos⁡u1ⅆu132+3⁢x8+∫cos⁡2⁢xⅆx2change,u1=2⁢u,u1=sin⁡u132+3⁢x8+∫cos⁡2⁢xⅆx2cos=sin⁡2⁢u32+3⁢x8+∫cos⁡2⁢xⅆx2revert=sin⁡4⁢x32+3⁢x8+∫cos⁡2⁢xⅆx2revert=sin⁡4⁢x32+3⁢x8+∫cos⁡uⅆu4change,u=2⁢x,u=sin⁡4⁢x32+3⁢x8+sin⁡u4cos=sin⁡4⁢x32+3⁢x8+sin⁡2⁢x4revert
Table 6.2.2(b) Annotated stepwise evaluation of ∫cos4x 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.
Table 6.2.2(c) applies the appropriate trig and algebra by means of the Maple command set.
Assign the unevaluated integral to a name
q1≔∫cos⁡x4ⅆx
∫cos⁡x4ⅆx
Apply the trig identity cos2x=1+cos2 x/2
q2≔simplifyalgsubscos2x=1+cos2 x/2,q1
14⁢∫cos⁡2⁢x2+2⁢cos⁡2⁢x+1ⅆx
Split the integral into three separate integrals
q3≔expandq2,cos2 x
14⁢∫cos⁡2⁢x2ⅆx+12⁢∫cos⁡2⁢xⅆx+14⁢∫1ⅆx
Apply the trig identity cos22 x=1+cos4 x/2
q4≔algsubscos22 x=1+cos4 x/2,q3
14⁢∫12⁢cos⁡4⁢x+12ⅆx+12⁢∫cos⁡2⁢xⅆx+14⁢∫1ⅆx
Split the first integral into two separate integrals, keeping cos2 x and cos4 x intact
expandq4,cos2 x,cos4 x
18⁢∫cos⁡4⁢xⅆx+38⁢∫1ⅆx+12⁢∫cos⁡2⁢xⅆx
Table 6.2.2(c) Command-based solution from first principles
Evaluation of the integrals in the last cell of Table 6.2.2(c) results in the solution given in Table 6.2.2(a).
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