100 Integrals (11-20) @blackpenredpen
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This post document the solutions to 100 integrals (Problem 11 to 20) solved by @blackpenredpen in his 5h++ marathon endeavor.
Contents:
- 11. \(\int \frac{\mathrm{sin}(x)}{\mathrm{sec}^{2019}(x)} \ dx\)
- 12. \(\int \frac{x\mathrm{sin}^{-1}(x)}{\sqrt{1 - x^{2}}} \ dx\)
- 13. \(\int \frac{2\mathrm{sin}(x)}{\mathrm{sin}(2x)}\ dx\)
- 14. \(\int \mathrm{cos}^{2}(2x) \ dx\)
- 15. \(\int \frac{1}{x^{3} + 1}\ dx\)
- 16. \(\int x \mathrm{sin}^{2}(x)\ dx\)
- 17. \(\int \Big(x + \frac{1}{x})^{2}\ dx\)
- 18. \(\int \frac{3}{x^{2} + 4x + 29} \ dx\)
- 19. \(\int \mathrm{cot}^{5} x\ dx\)
- 20. \(\int \frac{\mathrm{tan}(x)}{x^{4} - x^{2} + 1}\ dx\)
- See Also
- References
11. \(\int \frac{\mathrm{sin}(x)}{\mathrm{sec}^{2019}(x)} \ dx\)
\[\begin{aligned} \int \frac{\mathrm{sin}(x)}{\mathrm{sec}^{2019}(x)} \ dx &= \int \mathrm{cos}^{2019}(x)\mathrm{sin}(x)\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= \mathrm{cos}(x)\\ du &= -\mathrm{sin}(x)\ dx\\ dx &= \frac{du}{-\mathrm{sin}(x)}\ du \end{aligned}\] \[\begin{aligned} \int \mathrm{cos}^{2019}(x)\mathrm{sin}(x)\ dx &= \int u^{2019}\mathrm{sin}(x)\frac{du}{-\mathrm{sin}(x)}\ du\\ &= -\int u^{2019}\ du\\ &= -\frac{1}{2020}\mathrm{cos}^{2020}(x) + C \end{aligned}\]12. \(\int \frac{x\mathrm{sin}^{-1}(x)}{\sqrt{1 - x^{2}}} \ dx\)
\[\begin{aligned} \int \frac{x\mathrm{sin}^{-1}(x)}{\sqrt{1 - x^{2}}} \ dx \end{aligned}\]Integration by parts:
\[\begin{aligned} u &= \mathrm{sin}^{-1}(x)\\ dv &= \frac{x}{\sqrt{1 - x^{2}}} \ dx\\ v &= -\sqrt{1 - x^{2}}\\ du &= -\frac{1}{\sqrt{1 - x^{2}}}\\ \int u \ dv &= uv - \int v\ du\\ &= -\mathrm{sin}^{-1}(x)\sqrt{1 - x^{2}} - \int \sqrt{1 - x^{2}}\frac{1}{\sqrt{1 - x^{2}}}\ du\\ &= -\mathrm{sin}^{-1}(x)\sqrt{1 - x^{2}} - \int 1 \ dx\\ &= -\mathrm{sin}^{-1}(x)\sqrt{1 - x^{2}} - x + C \end{aligned}\] \[\begin{aligned} u &= \mathrm{sin}^{-1}(x)\\ dv &= \frac{x}{\sqrt{1 - x^{2}}} \ dx\\ v &= -\sqrt{1 - x^{2}}\\ du &= -\frac{1}{\sqrt{1 - x^{2}}}\\ \int u \ dv &= uv - \int v\ du\\ &= -\mathrm{sin}^{-1}(x)\sqrt{1 - x^{2}} - \int \sqrt{1 - x^{2}}\frac{1}{\sqrt{1 - x^{2}}}\ du\\ &= -\mathrm{sin}^{-1}(x)\sqrt{1 - x^{2}} - \int 1 \ dx\\ &= -\mathrm{sin}^{-1}(x)\sqrt{1 - x^{2}} - x + C &= 2xe^{x} - 2e^{x}\ dx \end{aligned}\]13. \(\int \frac{2sin(x)}{sin(2x)}\ dx\)
\[\begin{aligned} \int \frac{2\mathrm{sin}(x)}{\mathrm{sin}(2x)}\ dx \end{aligned}\]Double-Angle Formula:
\[\begin{aligned} \mathrm{sin}(2x) &= 2\mathrm{sin}(x)\mathrm{cos}(x) \end{aligned}\] \[\begin{aligned} \int \frac{2\mathrm{sin}(x)}{\mathrm{sin}(2x)}\ dx &= \int \frac{2\mathrm{sin}(x)}{2\mathrm{sin}(x)\mathrm{cos}(x)}\ dx\\ &= \int \mathrm{sec}(x) \ dx\\ &= int \frac{\mathrm{sec}(x)(\mathrm{sec}(x) + \mathrm{tan}(x))}{\mathrm{sec}(x) + \mathrm{tan}(x)}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= \mathrm{sec}(x) + \mathrm{tan}(x)\\ du &= \mathrm{sec}(x)\mathrm{tan}(x) + \mathrm{sec}^{2}(x)\ dx\\ dx &= \frac{1}{\mathrm{sec}(x)\mathrm{tan}(x) + \mathrm{sec}^{2}(x)}\ du \end{aligned}\] \[\begin{aligned} int \frac{\mathrm{sec}(x)(\mathrm{sec}(x) + \mathrm{tan}(x))}{\mathrm{sec}(x) + \mathrm{tan}(x)}\ dx &= \int \frac{1}{u}\ du\\ &= \mathrm{ln}|\mathrm{sec}(x) + \mathrm{tan}(x)| + C \end{aligned}\]14. \(\int \mathrm{cos}^{2}(2x) \ dx\)
\[\begin{aligned} \int \mathrm{cos}^{2}(2x) \ dx \end{aligned}\]Double-Angle Formula:
\[\begin{aligned} \mathrm{cos}^{2}(2x) &= \frac{1}{2}(1 + \mathrm{cos}(2x)) \end{aligned}\] \[\begin{aligned} \int \mathrm{cos}^{2}(2x) \ dx &= \frac{1}{2}\int 1 + \mathrm{cos}(4x)\ dx\\ &= \frac{1}{2}(x + \frac{1}{4}\mathrm{sin}(4x)) + C\\ &= \frac{1}{2}x + \frac{1}{8}\mathrm{sin}(4x) + C \end{aligned}\]15. \(\int \frac{1}{x^{3} + 1}\ dx\)
\[\begin{aligned} \int \frac{1}{x^{3} + 1}\ dx &= \int \frac{1}{(x+1)(x^{2} - x +1)}\ dx\\ &= \int \frac{A}{x + 1} + \frac{Bx + C}{x^{2} - x + 1}\\ &= \int \frac{\frac{1}{3}}{x + 1} + \frac{\frac{1}{3}x + \frac{2}{3}}{x^{2} - x + 1} \end{aligned}\]Partial Fractions:
\[\begin{aligned} \frac{1}{(x+1)(x^{2} - x +1)} &= \frac{A}{x + 1} + \frac{Bx + C}{x^{2} - x + 1}\\ &= \frac{\frac{1}{3}}{x + 1} + \frac{\frac{1}{3}x + \frac{2}{3}}{x^{2} - x + 1} \end{aligned}\] \[\begin{aligned} \int \frac{\frac{1}{3}}{x + 1} + \frac{\frac{1}{3}x + \frac{2}{3}}{x^{2} - x + 1} &= \frac{1}{3}\int \frac{1}{x+1}\ dx - \frac{1}{3}\int \frac{x + 2}{x^{2} - x + 1} \ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= x^{2} - x + 1\\ du &= 2x - 1\ dx\\ dx &= \frac{du}{2x - 1} \end{aligned}\] \[\begin{aligned} \frac{1}{3}\int \frac{1}{x+1}\ dx - \frac{1}{3}\int \frac{x + 2}{x^{2} - x + 1} \ dx &= \frac{1}{3}\int \frac{1}{x+1}\ dx - \frac{1}{3}\times\frac{1}{2}\int \frac{2(x + 2)}{x^{2} - x + 1} \ dx\\[5pt] &= \frac{1}{3}\int \frac{1}{x+1}\ dx - \frac{1}{6}\int \frac{2x -1 - 3}{x^{2} - x + 1} \ dx\\[5pt] &= \frac{1}{3}\int \frac{1}{x+1}\ dx - \frac{1}{6}\int \frac{2x -1}{x^{2} - x + 1} \ dx - \int \frac{3}{x^{2} - x + 1} \ dx\\[5pt] &= \frac{1}{3}\int \frac{1}{x+1}\ dx - \frac{1}{6}\int \frac{2x -1}{x^{2} - x + 1} \ dx - \int \frac{3}{x^{2} - x + \frac{1}{4} + \frac{3}{4}} \ dx\\[5pt] &= \frac{1}{3}\int \frac{1}{x+1}\ dx - \frac{1}{6}\int \frac{2x -1)}{x^{2} - x + 1} \ dx - \int \frac{3}{\Big(x - \frac{1}{2}\Big)^{2} + \Big(\frac{\sqrt{3}}{2}\Big)^{2}} \ dx\\[5pt] &= \frac{1}{3}\mathrm{ln}|x+1| - \frac{1}{6}\mathrm{ln}(x^{2} - x + 1) + \frac{1}{2}\times\frac{2}{\sqrt{3}}\mathrm{tan}^{-1}\Big(\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}\Big) + C\\[5pt] &= \frac{1}{3}\mathrm{ln}|x+1| - \frac{1}{6}\mathrm{ln}(x^{2} - x + 1) + \frac{1}{\sqrt{3}}\mathrm{tan}^{-1}\Big(\frac{2}{\sqrt{3}}x - \frac{1}{\sqrt{3}}\Big) + C \end{aligned}\]16. \(\int x \mathrm{sin}^{2}(x)\ dx\)
\[\begin{aligned} \int x \mathrm{sin}^{2}(x)\ dx \end{aligned}\]Double-Angle Formula:
\[\begin{aligned} \mathrm{sin}^{2}(x) = \frac{1}{2}(1 - \mathrm{cos}(2x)) \end{aligned}\] \[\begin{aligned} \int x \mathrm{sin}^{2}(x)\ dx &= \frac{1}{2}\int x(1 - \mathrm{cos}(2(x)))\ dx\\ &= \frac{1}{2}\Big(\int x\ dx - \int x\mathrm{cos}(2x)\Big)\ dx \end{aligned}\]Integration by parts:
\[\begin{aligned} u &= x\\ dv &= \mathrm{cos}(2x)\ dx\\ v &= \frac{1}{2}\mathrm{sin}(2x)\\ du &= dx\\ \int u \ dv &= uv - \int v\ du\\ &= \frac{1}{2}x\mathrm{sin}(2x) - \frac{1}{2}\int \mathrm{sin}(2x)\ dx\\ &= \frac{1}{2}x\mathrm{sin}(2x) + \frac{1}{4}\mathrm{cos}(2x)\ dx \end{aligned}\] \[\begin{aligned} \frac{1}{2}\Big(\int x\ dx - \int x\mathrm{cos}(2x)\Big)\ dx &= \frac{1}{2}\Big(\frac{1}{2}x^{2} - \frac{1}{2}x\mathrm{sin}(2x) - \frac{1}{4}\mathrm{cos}(2x)\Big) + C\\ &= \frac{1}{4}x^{2} - \frac{1}{4}x\mathrm{sin}(2x) - \frac{1}{8}\mathrm{cos}(2x) + C \end{aligned}\]17. \(\int \Big(x + \frac{1}{x})^{2}\ dx\)
\[\begin{aligned} \int \Big(x + \frac{1}{x}\Big)^{2}\ dx &= \int x^{2} + 2 + x^{-2}\ dx\\ &= \frac{1}{3}x^{3} + 2x - \frac{1}{x} + C \end{aligned}\]18. \(\int \frac{3}{x^{2} + 4x + 29} \ dx\)
\[\begin{aligned} \int \frac{3}{x^{2} + 4x + 29} \ dx &= \int \frac{3}{x^{2} + 4x + 4 + 25}\ dx &= \int \frac{3}{(x+2)^{2} + 5^{2}}\ dx\\ &= \frac{3}{5}\mathrm{tan}^{-1}\Big(\frac{x+2}{5}\Big) + C \end{aligned}\]19. \(\int \mathrm{cot}^{5} x\ dx\)
\[\begin{aligned} \int \mathrm{cot}^{5} x\ dx &= \int \frac{\mathrm{cos}^{5}(x)}{\mathrm{sin}^{5}(x)}\ dx\\ &= \int \frac{\mathrm{cos}^{4}(x)\mathrm{cos}(x)}{\mathrm{sin}^{5}(x)}\ dx\\ &= \int \frac{(1 - \mathrm{sin}^{2}(x))^{2}\mathrm{cos}(x)}{\mathrm{sin}^{5}(x)}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= \mathrm{sin}(x)\\ du &= \mathrm{cos}(x)\ dx\\ dx &= \frac{du}{\mathrm{cos}(x)} \end{aligned}\] \[\begin{aligned} \int \frac{(1 - \mathrm{sin}^{2}(x))^{2}\mathrm{cos}(x)}{\mathrm{sin}^{5}(x)}\ dx &= \int \frac{(1 - u^{2})^{2}}{u^{5}}\ du\\ &= \int \frac{1 - 2u^{2} + u^{4}}{u^{5}}\ du\\ &= \int u^{-5} - 2u^{-3} + \frac{1}{u}\ du\\ &= -\frac{1}{4}u^{-4} - \frac{2}{-2}u^{-2} + \mathrm{ln}|u| + C\\ &= -\frac{1}{4}csc^{4}(x) + csc^{2}(x) + \mathrm{ln}|\mathrm{sin}(x)| + C \end{aligned}\]20. \(\int \frac{\mathrm{tan}(x)}{x^{4} - x^{2} + 1}\ dx\)
\[\begin{aligned} \int \frac{\mathrm{tan}(x)}{x^{4} - x^{2} + 1}\ dx \end{aligned}\]Because \(\mathrm{tan}(x)\) is an odd function, and \(x^{4} - x^{2} + 1\) is an even function (see reference):
\[\begin{aligned} \frac{\text{odd function}}{\text{even function}} &= \text{odd function} \end{aligned}\] \[\begin{aligned} \int \frac{\mathrm{tan}(x)}{x^{4} - x^{2} + 1}\ dx &= 0 \end{aligned}\]