100 Integrals (51-60) @blackpenredpen
Read PostThis post document the solutions to 100 integrals (Problem 41 to 50) solved by @blackpenredpen in his 5h++ marathon endeavor.
Contents:
- 51. \(\int \mathrm{sec}^{6}(x)\ dx\)
- 52. \(\int \frac{1}{(5x - 2)^{4}}\ dx\)
- 53. \(\int \mathrm{ln}(1 + x^{2})\ dx\)
- 54. \(\int \frac{1}{x^{4} + x}\ dx\)
- 55. \(\int \frac{1 - \mathrm{tan}(x)}{1 + \mathrm{tan}(x)}\ dx\)
- 56. \(\int x \mathrm{sec}(x) \mathrm{tan}(x)\ dx\)
- 57. \(\int \mathrm{sec}^{-1}(x)\ dx\)
- 58. \(\int\frac{1 - \mathrm{cos}(x)}{1 + \mathrm{cos}(x)}\ dx\)
- 59. \(\int x^{2}\sqrt{x + 4}\ dx\)
- 60. \(\int_{-1}^{1} \sqrt{4 - x^{2}}\ dx\)
- See Also
- References
60. \(\int_{-1}^{1} \sqrt{4 - x^{2}}\ dx\)
\[\begin{aligned} \int_{-1}^{1} \sqrt{4 - x^{2}}\ dx \end{aligned}\]51. \(\int \mathrm{sec}^{6}(x)\ dx\)
\[\begin{aligned} \int \mathrm{sec}^{6}(x)\ dx &= \int (\mathrm{tan}^{2}(x) + 1)^{2}\mathrm{sec}^{2}(x)\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= \mathrm{tan}(x)\\ du &= \mathrm{sec}^{2}(x) \ dx\\ dx &= \frac{du}{\mathrm{sec}^{2}(x)} \end{aligned}\] \[\begin{aligned} \int (\mathrm{tan}^{2}(x) + 1)^{2}\mathrm{sec}^{2}(x)\ dx &= \int (u^{2} + 1)^{2}\mathrm{sec}^{2}(x)\frac{du}{\mathrm{sec}^{2}(x)}\\ &= \int (u^{2} + 1)^{2}\ du\\ &= \int u^{4} + 2u^{2} + 1\ du\\ &= \frac{1}{5}u^{5} + \frac{2}{3}u^{3} + u + C\\ &= \frac{1}{5}\mathrm{tan}^{5}(x) + \frac{2}{3}\mathrm{tan}^{3}(x) + \mathrm{tan}(x) + C\\ \end{aligned}\]52. \(\int \frac{1}{(5x - 2)^{4}}\ dx\)
\[\begin{aligned} \int \frac{1}{(5x - 2)^{4}}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= 5x - 2\\ du &= 5\ dx\\ dx &= \frac{du}{5} \end{aligned}\] \[\begin{aligned} \int \frac{1}{(5x - 2)^{4}}\ dx &= \int \frac{1}{u^{4}} \times \frac{du}{5}\\ &= \frac{1}{5}\int \frac{1}{u^{4}}\ du\\ &= -\frac{1}{15u^{3}} + C\\ &= -\frac{1}{15(5x - 2)^{3}} + C\\ \end{aligned}\]53. \(\int \mathrm{ln}(1 + x^{2})\ dx\)
\[\begin{aligned} \int \mathrm{ln}(1 + x^{2})\ dx \end{aligned}\]Integration by parts:
\[\begin{aligned} u &= \mathrm{ln}(1 + x)^{2}\\ dv &= dx\\ v &= x\\ du &= -\frac{2x}{1 + x^{2}}\ dx\\ \int u \ dv &= uv - \int v\ du\\ &= \mathrm{ln}(1 + x)^{2} - \int \frac{2x^{2}}{1 + x^{2}}\ dx\\ &= x\mathrm{ln}(1 + x^{2}) -2\int \frac{1 + x^{2} - 1}{1 + x^{2}}\ dx\\ &= x\mathrm{ln}(1 + x^{2}) -2\int \ dx + 2\int \frac{1}{1 + x^{2}}\ dx\\ &= x\mathrm{ln}(1 + x^{2}) -2x + 2\mathrm{tan}^{-1}(x) + C \end{aligned}\]54. \(\int \frac{1}{x^{4} + x}\ dx\)
\[\begin{aligned} \int \frac{1}{x^{4} + x}\ dx &= \int \frac{1}{x^{4}(1 + x^{-3})}\ dx\\ &= \int \frac{x^{-4}}{1 + x^{-3}}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= 1 + x^{-3} du &= -3x^{-4}\ dx\\ dx &= -\frac{du}{3x^{-4}} \end{aligned}\] \[\begin{aligned} \int \frac{x^{-4}}{1 + x^{-3}}\ dx &= \int \frac{1}{u}\frac{du}{-3x^{-4}}\\ &= -\frac{1}{3}\mathrm{ln}|u| + C\\ &= -\frac{1}{3}\mathrm{ln}|1 + x^{-3}| + C\\ \end{aligned}\]55. \(\int \frac{1 - \mathrm{tan}(x)}{1 + \mathrm{tan}(x)}\ dx\)
\[\begin{aligned} \int \frac{1 - \mathrm{tan}(x)}{1 + \mathrm{tan}(x)}\ dx &= \int \frac{1 - \frac{\mathrm{sin}(x)}{\mathrm{cos}(x)}}{1 + \frac{\mathrm{sin}(x)}{\mathrm{cos}(x)}}\ dx\\ &= \int \frac{\frac{\mathrm{cos}(x) - \mathrm{sin}(x)}{\mathrm{cos}(x)}}{\frac{\mathrm{cos}(x) + \mathrm{sin}(x)}{\mathrm{cos}(x)}}\ dx\\ &= \int \frac{\mathrm{cos}(x) - \mathrm{sin}(x)}{\mathrm{cos}(x) + \mathrm{sin}(x)}\ dx\\ \end{aligned}\] \[\begin{aligned} u &= \mathrm{cos}(x) + \mathrm{sin}(x)\\ du &= -\mathrm{sin}(x) + \mathrm{cos}(x)\ dx\\ dx &= \frac{du}{\mathrm{cos}(x) - \mathrm{sin}(x)} \end{aligned}\] \[\begin{aligned} \int \frac{\mathrm{cos}(x) - \mathrm{sin}(x)}{\mathrm{cos}(x) + \mathrm{sin}(x)}\ dx &= \int \frac{\mathrm{cos}(x) - \mathrm{sin}(x)}{u} \frac{du}{\mathrm{cos}(x) - \mathrm{sin}(x)}\\ &= \int \frac{1}{u}\ du\\ &= \mathrm{ln}|u| + C\\ &= \mathrm{ln}|\mathrm{cos}(x) + \mathrm{sin}(x)| + C \end{aligned}\]56. \(\int x \mathrm{sec}(x) \mathrm{tan}(x)\ dx\)
\[\begin{aligned} \int x \mathrm{sec}(x) \mathrm{tan}(x)\ dx \end{aligned}\]Integration by parts:
\[\begin{aligned} u &= x\\ dv &= \mathrm{sec}(x)\mathrm{tan}(x)\ dx\\ v &= \mathrm{sec}(x)\\ du &= dx\\ \int u \ dv &= uv - \int v\ du\\ &= x \mathrm{sec}(x) - \int \mathrm{sec}(x)\ dx\\ &= x \mathrm{sec}(x) - \mathrm{ln}|\mathrm{sec}(x) + \mathrm{tan}(x)| + C \end{aligned}\] \[\begin{aligned} \int x \mathrm{sec}(x) \mathrm{tan}(x)\ dx &= x \mathrm{sec}(x) - \mathrm{ln}|\mathrm{sec}(x) + \mathrm{tan}(x)| + C \end{aligned}\]57. \(\int \mathrm{sec}^{-1}(x)\ dx\)
\[\begin{aligned} \int \mathrm{sec}^{-1}(x)\ dx \end{aligned}\]Integration by parts:
\[\begin{aligned} u &= \mathrm{sec}^{-1}(x)\\ dv &= dx\\ v &= x\\ du &= \frac{1}{x\sqrt{x^{2}-1}}\\ \int u \ dv &= uv - \int v\ du\\ &= x \mathrm{sec}^{-1}(x) - \int \frac{1}{\sqrt{x^{2}-1}}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} x &= \mathrm{sec}(\theta)\\ dx &= \mathrm{sec}(\theta)\mathrm{tan}(\theta)\ d\theta \end{aligned}\] \[\begin{aligned} \int \frac{1}{\sqrt{x^{2}-1}}\ dx &= \int \frac{1}{\sqrt{\mathrm{sec}^{2}(\theta) - 1}}\mathrm{sec}(\theta)\mathrm{tan}(\theta)\ d\theta\\ &= \int \frac{1}{\sqrt{\mathrm{tan}^{2}(\theta)}}\mathrm{sec}(\theta)\mathrm{tan}(\theta)\ d\theta\\ &= \int \mathrm{sec}(\theta)\ d\theta\\ &= \mathrm{ln}|\mathrm{sec}(\theta) + \mathrm{tan}(\theta)| + C\\ &= \mathrm{ln}|x + \sqrt{x^{2} - 1}| + C \end{aligned}\] \[\begin{aligned} x \mathrm{sec}^{-1}(x) - \int \frac{1}{\sqrt{x^{2}-1}}\ dx &= x \mathrm{sec}^{-1}(x) - \mathrm{ln}|x + \sqrt{x^{2} - 1}| + C \end{aligned}\]58. \(\int\frac{1 - \mathrm{cos}(x)}{1 + \mathrm{cos}(x)}\ dx\)
\[\begin{aligned} \int\frac{1 - \mathrm{cos}(x)}{1 + \mathrm{cos}(x)}\ dx \end{aligned}\]Trig Identities:
\[\begin{aligned} \mathrm{cos}^{2}(\theta) &= \frac{1}{2}(1 + \mathrm{cos}(2\theta))\\ \mathrm{sin}^{2}(\theta) &= \frac{1}{2}(1 - \mathrm{cos}(2\theta))\\ \end{aligned}\] \[\begin{aligned} \int\frac{1 - \mathrm{cos}(x)}{1 + \mathrm{cos}(x)}\ dx &= \int \frac{2 \mathrm{sin}^{2}\Big(\frac{x}{2}\Big)}{2 \mathrm{cos}^{2}\Big(\frac{x}{2}\Big)}\\ &= \int \mathrm{tan}^{2}\Big(\frac{x}{2}\Big)\ dx\\ &= \int sec^{2}\Big(\frac{x}{2}\Big) - 1 \ dx\\ &= 2 \mathrm{tan}\Big(\frac{x}{2}\Big) - x + C \end{aligned}\]59. \(\int x^{2}\sqrt{x + 4}\ dx\)
\[\begin{aligned} \int x^{2}\sqrt{x + 4}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= x + 4\\ du &= dx \end{aligned}\] \[\begin{aligned} \int x^{2}\sqrt{x + 4}\ dx &= \int (u - 4)^{2}\sqrt{u}\ du\\ &= \int (u^{2} - 8u + 16)u^{\frac{1}{2}}\ du\\ &= \int u^{\frac{5}{2}} - 8u^{\frac{3}{2}} + 16u^{\frac{1}{2}}\\ &= \frac{2}{7}u^{\frac{7}{2}} - \frac{2}{5}8u^{\frac{5}{2}} + \frac{2}{3}16u^{\frac{3}{2}} + C\\ &= \frac{2}{7}(x + 4)^{\frac{7}{2}} - \frac{15}{5}(x + 4)^{\frac{5}{2}} + \frac{32}{3}(x + 4)^{\frac{3}{2}} + C \end{aligned}\]