100 Integrals (41-50) @blackpenredpen
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This post document the solutions to 100 integrals (Problem 41 to 50) solved by @blackpenredpen in his 5h++ marathon endeavor.
Contents:
- 41. \(\int \mathrm{sinh}(x)\ dx\)
- 42. \(\int \mathrm{sinh}^{2}(x)\ dx\)
- 43. \(\int \mathrm{sinh}^{3}(x)\ dx\)
- 44. \(\int \frac{1}{\sqrt{x^{2} + 1}}\ dx\)
- 45. \(\int \mathrm{ln}(x + \sqrt{1 + x^{2}})\ dx\)
- 46. \(\int \mathrm{tanh}(x)\ dx\)
- 47. \(\int \mathrm{sech}(x)\ dx\)
- 48. \(\int \mathrm{tanh}^{-1}(x)\ dx\)
- 49. \(\int \sqrt{\mathrm{tanh}(x)}\ dx\)
- 50. \(\int_{0}^{5} \lfloor x \rfloor \ dx\)
- See Also
- References
41. \(\int \mathrm{sinh}(x)\ dx\)
\[\begin{aligned} \int \mathrm{sinh}(x)\ dx &= \int \frac{e^{x} - e^{-x}}{2}\ dx\\ &= \frac{1}{2}\int e^{x} + e^{-x}\ dx\\ &= \mathrm{cosh}(x) + C \end{aligned}\]42. \(\int \mathrm{sinh}^{2}(x)\ dx\)
\[\begin{aligned} \int \mathrm{sinh}^{2}(x)\ dx &= \int \Big( \frac{e^{x} - e^{-x}}{2} \Big)^{2}\ dx\\ &= \int \frac{1}{4}(e^{2x} - 2 + e^{-2x})\ dx\\ &= \int \frac{1}{4}(-2) + \frac{1}{2\times 2}(e^{2x} + e^{-2x})\ dx\\ &= \int -\frac{1}{2} + \frac{1}{2}\mathrm{cosh}(2x)\ dx\\ &= \frac{1}{2}\int -1 + \mathrm{cosh}(2x)\ dx\\ &= \frac{1}{2}(-x + \frac{1}{2}\mathrm{sinh}(2x)) + C\\ &= -\frac{1}{2}x + \frac{1}{4}\mathrm{sinh}(2x) + C \end{aligned}\]43. \(\int \mathrm{sinh}^{3}(x)\ dx\)
\[\begin{aligned} \int \mathrm{sinh}^{3}(x)\ dx &= \int \mathrm{sinh}^{2}(x)\times \mathrm{sinh}(x)\ dx\\ &= \int (\mathrm{cosh}^{2}(x) - 1) \mathrm{sinh}(x)\ dx\\ \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= \mathrm{cosh}(x)\\ du &= \mathrm{sinh}(x)\ dx\\ dx &= \frac{1}{\mathrm{sinh}(x)}\ du \end{aligned}\] \[\begin{aligned} \int (\mathrm{cosh}^{2}(x) - 1) \mathrm{sinh}(x)\ dx &= \int (u^{2}(x) - 1) \mathrm{sinh}(x)\frac{1}{\mathrm{sinh}(x)}\ du\\ &= \int u^{2}(x) - 1 \ du\\ &= \frac{1}{3}u^{3} - u + C\\ &= \frac{1}{3}\mathrm{cosh}(x)^{3} - \mathrm{cosh}(x) + C \end{aligned}\]44. \(\int \frac{1}{\sqrt{x^{2} + 1}}\ dx\)
\[\begin{aligned} \int \frac{1}{\sqrt{x^{2} + 1}}\ dx &= \int \frac{1}{\sqrt{1 + x^{2}}}\ dx\\ &= \mathrm{sinh}^{-1}(x) + C\\ &= \mathrm{ln}(x + \sqrt{1 + x^{2}}) + C \end{aligned}\]45. \(\int \mathrm{ln}(x + \sqrt{1 + x^{2}})\ dx\)
\[\begin{aligned} \int \mathrm{ln}(x + \sqrt{1 + x^{2}})\ dx &= \int \mathrm{sinh}^{-1}(x)\ dx \end{aligned}\]Integration by parts:
\[\begin{aligned} u &= \mathrm{sinh}^{-1}(x)\\ dv &= dx\\ v &= x\\ du &= -\frac{1}{\sqrt{1 + x^{2}}}\ dx\\ \int u \ dv &= uv - \int v\ du\\ &= x\mathrm{sinh}^{-1}(x) - \int x \times -\frac{1}{\sqrt{1 + x^{2}}}\ dx\\ &= x\mathrm{sinh}^{-1}(x) + \int \frac{x}{\sqrt{1 + x^{2}}}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= 1 + x^{2}\\ du &= 2x\ dx\\ dx &= \frac{1}{2x}\ du \end{aligned}\] \[\begin{aligned} x\mathrm{sinh}^{-1}(x) + \int \frac{x}{\sqrt{1 + x^{2}}}\ dx &= x\mathrm{sinh}^{-1}(x) + \int \frac{x}{\sqrt{u}} \times \frac{du}{2x}\\ &= x\mathrm{sinh}^{-1}(x) + \int \frac{1}{\sqrt{u}} \ du\\ &= x\mathrm{sinh}^{-1}(x) - \int \frac{1}{2\sqrt{u}} \ du\\ &= x\mathrm{sinh}^{-1}(x) - \frac{2}{2}\sqrt{u} \ du\\ &= x\mathrm{sinh}^{-1}(x) - \sqrt{1 + x^{2}} + C \end{aligned}\]46. \(\int \mathrm{tanh}(x)\ dx\)
\[\begin{aligned} \int \mathrm{tanh}(x)\ dx &= \int \frac{\mathrm{sinh}(x)}{\mathrm{cosh}(x)}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= \mathrm{cosh}(x)\\ du &= \mathrm{sinh}(x)\ dx\\ dx &= \frac{1}{\mathrm{sinh}(x)}\ du \end{aligned}\] \[\begin{aligned} \int \frac{\mathrm{sinh}(x)}{\mathrm{cosh}(x)}\ dx &= \int \frac{\mathrm{sinh}(x)}{u}\frac{1}{\mathrm{sinh}(x)}\ du\\ &= \mathrm{ln}|u| + C\\ &= \mathrm{ln}|\mathrm{cosh}(x)| + C \end{aligned}\]47. \(\int \mathrm{sech}(x)\ dx\)
\[\begin{aligned} \int \mathrm{sech}(x)\ dx &= \frac{1}{\mathrm{cosh}(x)}\ dx\\ &= \frac{\mathrm{cosh}(x)}{\mathrm{cosh}^{2}(x)}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= \mathrm{sinh}(x)\\ du &= \mathrm{cosh}(x)\ dx\\ dx &= \frac{1}{\mathrm{cosh}(x)}\ du \end{aligned}\] \[\begin{aligned} \frac{\mathrm{cosh}(x)}{\mathrm{cosh}^{2}(x)}\ dx &= \int \frac{\mathrm{cosh}(x)}{1 + u^{2}}\frac{1}{\mathrm{\mathrm{cosh}}(x)}\ du\\ &= \int \frac{1}{1+u^{2}}\ du\\ &= \mathrm{tan}^{-1}(u) + C\\ &= \mathrm{tan}^{-1}(\mathrm{sinh}(x)) + C\\ \end{aligned}\]48. \(\int \mathrm{tanh}^{-1}(x)\ dx\)
\[\begin{aligned} \int \mathrm{tanh}^{-1}(x)\ dx \end{aligned}\]Integration by parts:
\[\begin{aligned} u &= \mathrm{tanh}^{-1}(x)\\ dv &= dx\\ v &= x\\ du &= -\frac{1}{1 - x^{2}}\ dx\\ \int u \ dv &= uv - \int v\ du\\ &= x\mathrm{tanh}^{-1}(x) + \int \frac{x}{1 - x^{2}}\ dx\\ \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= 1 - x^{2}\\ du &= -2x \ dx\\ dx &= -\frac{du}{2x} \end{aligned}\] \[\begin{aligned} x\mathrm{tanh}^{-1}(x) + \int \frac{x}{\sqrt{1 - x^{2}}}\ dx &= x\mathrm{tanh}^{-1}(x) - \int \frac{x}{u}\frac{du}{2x}\\ &= x\mathrm{tanh}^{-1}(x) + \int \frac{1}{2u}\ du\\ &= x\mathrm{tanh}^{-1}(x) + \int \frac{1}{2}\mathrm{ln}|1 - x^{2}|\ du\\ \end{aligned}\]49. \(\int \sqrt{\mathrm{tanh}(x)}\ dx\)
\[\begin{aligned} \int \sqrt{\mathrm{tanh}(x)}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= \sqrt{\mathrm{tanh}(x)}\\ u^{2} &= \mathrm{tanh}(x)\\ x &= \mathrm{tanh}^{-1}(u^{2})\\ dx &= \frac{2u}{1 - (u^{2})^{2}}\ du\\ &= \frac{2u}{1 - u^{4}}\ du \end{aligned}\] \[\begin{aligned} \int \sqrt{\mathrm{tanh}(x)}\ dx &= \int u \times \frac{2u}{1 - u^{4}}\ du\\ &= \int \frac{2u^{2}}{1 - u^{4}}\ du\\ &= \int \frac{2u^{2}}{(1 - u^{2})(1 + u^{2})}\ du\\ &= \int \frac{u^{2} - 1 + u^{2} + 1}{(1 - u^{2})(1 + u^{2})}\ du\\ &= \int \frac{-(1 - u^{2})}{(1 - u^{2})(1 + u^{2})} + \frac{1 + u^{2}}{(1 - u^{2})(1 + u^{2})}\ du\\ &= \int \frac{-1}{1 + u^{2}} + \frac{1}{1 - u^{2}}\ du\\ &= \mathrm{tan}^{-1}(u) + \mathrm{tanh}^{-1}(u) + C\\ &= \mathrm{tan}^{-1}\Big(\sqrt{\mathrm{tanh}(x)}\Big) + \mathrm{tanh}^{-1}\Big(\sqrt{\mathrm{tanh}(x)}\Big) + C\\ \end{aligned}\]