Chapter 6: Techniques of Integration
Section 6.1: Integration by Parts
Use integration by parts to evaluate the indefinite integral ∫x2sinx ⅆx.
The given integral is readily evaluated with parts integration applied twice, the first time with u=x2; and the second, with u=x. The complete solution appears below.
=x2−cosx−∫2 x −cosx ⅆx
u=x2⇒du=2 x dx
=−x2cosx+2∫x cosx ⅆx
=−x2cosx+2x sinx−∫sinx ⅆx
=−x2cosx+2x sinx−− cosx
=−x2cosx+2 x sinx+2 cosx
The following annotated stepwise solution is obtained via the Context Panel. Once again, note that when Maple indicates the Parts rule has been applied, both u and v are given (in that order). An equivalent solution can also be obtained interactively via the
Tools≻Load Package: Student Calculus 1
Control-drag the given integral.
Context Panel: Student Calculus1≻All Solution Steps
∫x2sinx ⅆx→show solution stepsIntegration Steps∫x2⁢sin⁡xⅆx▫1. Apply integration by Parts◦Recall the definition of the Parts rule∫uⅆv=v⁢u−∫vⅆu◦First partu=x2◦Second part=sin⁡x◦Differentiate first part=ⅆⅆxx2=2⁢x◦Integrate second partv=∫sin⁡xⅆxv=−cos⁡x∫x2⁢sin⁡xⅆx=−x2⁢cos⁡x−∫−2⁢cos⁡x⁢xⅆxThis gives:−x2⁢cos⁡x−∫−2⁢cos⁡x⁢xⅆx▫2. Apply the constant multiple rule to the term ∫−2⁢cos⁡x⁢xⅆx◦Recall the definition of the constant multiple rule∫⁢f⁡xⅆx=⁢∫f⁡xⅆx◦This means:∫−2⁢cos⁡x⁢xⅆx=−2⁢∫cos⁡x⁢xⅆxWe can rewrite the integral as:−x2⁢cos⁡x+2⁢∫cos⁡x⁢xⅆx▫3. Apply integration by Parts◦Recall the definition of the Parts rule∫uⅆv=v⁢u−∫vⅆu◦First partu=x◦Second part=cos⁡x◦Differentiate first part=ⅆⅆxx=1◦Integrate second partv=∫cos⁡xⅆxv=sin⁡x∫cos⁡x⁢xⅆx=x⁢sin⁡x−∫sin⁡xⅆxThis gives:−x2⁢cos⁡x+2⁢x⁢sin⁡x−2⁢∫sin⁡xⅆx▫4. Evaluate the integral of sin(x)◦Recall the definition of the sin rule∫sin⁡xⅆx=−cos⁡xThis gives:−x2⁢cos⁡x+2⁢x⁢sin⁡x+2⁢cos⁡x
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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