100 Integrals (31-40) @blackpenredpen
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This post document the solutions to 100 integrals (Problem 31 to 40) solved by @blackpenredpen in his 5h++ marathon endeavor.
Contents:
- 31. \(\int \frac{1}{\sqrt{x - x^{\frac{3}{2}}}} \ dx\)
- 32. \(\int \frac{1}{\sqrt{x - x^{2}}}\ dx\)
- 33. \(\int e^{2\mathrm{ln}(x)}\ dx\)
- 34. \(\int \frac{\mathrm{ln}(x)}{\sqrt{x}}\ dx\)
- 35. \(\int \frac{1}{e^{x} + e^{-x}}\ dx\)
- 36. \(\int \mathrm{log}_{2}(x)\ dx\)
- 37. \(\int x^{3}\mathrm{sin}(2x)\ dx\)
- 38. \(\int x^{2}\sqrt[3]{1 + x^{3}} \ dx\)
- 39. \(\int \frac{1}{(x^{2} + 4)^{2}}\ dx\)
- 40. \(\int_{1}^{2} \sqrt{x^{2} - 1} \ dx\)
- See Also
- References
31. \(\int \frac{1}{\sqrt{x - x^{\frac{3}{2}}}} \ dx\)
\[\begin{aligned} \int \frac{1}{\sqrt{x - x^{\frac{3}{2}}}} \ dx &= \int \frac{1}{\sqrt{x}\sqrt{1 - x^{\frac{1}{2}}}} \ dx\\ &= \int \frac{1}{\sqrt{x}\sqrt{1 - \sqrt{x}}} \ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= 1 - \sqrt{x}\\ du &= -\frac{1}{2\sqrt{x}}\ dx\\ dx &= -2\sqrt{x}\ du \end{aligned}\] \[\begin{aligned} \int \frac{1}{\sqrt{x}\sqrt{1 - \sqrt{x}}} \ dx &= -\int \frac{2\sqrt{x}}{\sqrt{x}\sqrt{u}}\ du\\ &= -2\int u^{-\frac{1}{2}}\ du\\ &= -2\times 2u^{-\frac{1}{2} + 1}\\ &= -4 u^{\frac{1}{2}} + C\\ &= -4\sqrt{1 - \sqrt{x}} + C \end{aligned}\]32. \(\int \frac{1}{\sqrt{x - x^{2}}}\ dx\)
\[\begin{aligned} \int \frac{1}{\sqrt{x - x^{2}}}\ dx &= \int \frac{1}{\sqrt{x}\sqrt{1 - x}}\ dx\\ &= \int \frac{1}{\sqrt{x}\sqrt{1 - \sqrt{x}^{2}}}\ dx\\ \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= \sqrt{x}\\ du &= \frac{1}{2\sqrt{x}}\ dx\\ dx &= 2\sqrt{x}\ du \end{aligned}\] \[\begin{aligned} \int \frac{1}{\sqrt{x}\sqrt{1 - \sqrt{x}^{2}}}\ dx &= \int \frac{1}{\sqrt{x}\sqrt{1 - u^{2}}}\times 2\sqrt{x}\ du\\ &= \int \frac{2}{\sqrt{1 - u^{2}}}\ du\\ &= 2 \mathrm{sin}(u) + C\\ &= 2 \mathrm{sin}(\sqrt{x}) + C \end{aligned}\]33. \(\int e^{2\mathrm{ln}(x)}\ dx\)
\[\begin{aligned} \int e^{2\mathrm{ln}(x)}\ dx &= \int e{^\mathrm{ln}(x^{2})}\ dx &= \int x^{2}\ dx\\ &= \frac{1}{3}x^{3} + C \end{aligned}\]34. \(\int \frac{\mathrm{ln}(x)}{\sqrt{x}}\ dx\)
\[\begin{aligned} \int \frac{\mathrm{ln}(x)}{\sqrt{x}}\ dx \end{aligned}\]Integration by parts:
\[\begin{aligned} u &= \mathrm{ln}(x)\\ dv &= x^{\frac{1}{2}}\ dx\\ v &= 2\sqrt{x}\\ du &= \frac{1}{x}\\ \int u \ dv &= uv - \int v\ du\\ &= 2\sqrt{x}\mathrm{ln}(x) - 2\int \frac{\sqrt{x}}{x}\ dx\\ &= 2\sqrt{x}\mathrm{ln}(x) - 2\int \frac{1}{\sqrt{x}}\ dx \end{aligned}\] \[\begin{aligned} \int \frac{\mathrm{ln}(x)}{\sqrt{x}}\ dx &= 2\sqrt{x}\mathrm{ln}(x) - 2\int \frac{1}{\sqrt{x}}\ dx\\ &= 2\sqrt{x}\mathrm{ln}(x) - 2\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C\\ &= 2\sqrt{x}\mathrm{ln}(x) - 4x^{\frac{1}{2}} + C \end{aligned}\]35. \(\int \frac{1}{e^{x} + e^{-x}}\ dx\)
\[\begin{aligned} \int \frac{1}{e^{x} + e^{-x}}\ dx &= \int \frac{1 \times e^{x}}{(e^{x} + e^{-x})\times e^{x}}\ dx\\ &= \int \frac{e^{x}}{e^{2x} + 1}\ dx\\ &= \int \frac{e^{x}}{(e^{x})^{2} + 1}\ dx\\ \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= e^{x}\\ du &= e^{x}\ dx\\ dx &= \frac{du}{e^{x}} \end{aligned}\] \[\begin{aligned} \int \frac{e^{x}}{(e^{x})^{2} + 1}\ dx &= \int \frac{e^{x}}{u^{2} + 1}\frac{du}{e^{x}}\\ &= \int \frac{1}{u^{2} + 1}\ du\\ &= \mathrm{tan}^{-1}(u) + C\\ &= \mathrm{tan}^{-1}(e^{x}) + C \end{aligned}\]36. \(\int \mathrm{log}_{2}(x)\ dx\)
\[\begin{aligned} \int \mathrm{log}_{2}(x)\ dx &= \int \frac{\mathrm{ln}(x)}{\mathrm{ln}(2)}\ dx \end{aligned}\]Integration by parts:
\[\begin{aligned} u &= \mathrm{ln}(x)\\ dv &= dx\\ v &= x\\ du &= \frac{1}{x}\ dx\\ \int u \ dv &= uv - \int v\ du\\ &= x\mathrm{ln}(x) - \int x\times\frac{1}{x}\ dx\\ &= x\mathrm{ln}(x) - \int 1\ dx\\ &= x\mathrm{ln}(x) - x \end{aligned}\] \[\begin{aligned} \int \frac{\mathrm{ln}(x)}{\mathrm{ln}(2)}\ dx &= \frac{1}{\mathrm{ln}(2)}(x\mathrm{ln}(x) - x) + C\\ &= \frac{x\mathrm{ln}(x)}{\mathrm{ln}(2)} - \frac{x}{\mathrm{ln}(2)} + C\\ &= x \mathrm{log}_{2}(x) - \frac{x}{\mathrm{ln}(2)} + C\\ \end{aligned}\]37. \(\int x^{3}\mathrm{sin}(2x)\ dx\)
\[\begin{aligned} \int x^{3}\mathrm{sin}(2x)\ dx \end{aligned}\]Integration by parts:
Differentiate:
\[\begin{aligned} +x^{3}\\ -3x^{2}\\ +6x\\ -6 \end{aligned}\]Integrate:
\[\begin{aligned} \mathrm{sin}(2x)\\ -\frac{1}{2}\mathrm{cos}(2x)\\ -\frac{1}{4}\mathrm{sin}(2x)\\ \frac{1}{8}\mathrm{cos}(2x)\\ \frac{1}{16}\mathrm{sin}(2x)\\ \end{aligned}\] \[\begin{aligned} \int x^{3}\mathrm{sin}(2x)\ dx &= x^{3} \times -\frac{1}{2}\mathrm{cos}(2x) - 3x^{2}\times -\frac{1}{4}\mathrm{sin}(2x) + 6x\times \frac{1}{8}\mathrm{cos}(2x) - 6 \times \frac{1}{16}\mathrm{sin}(2x) + C\\ &= -\frac{1}{2}x^{3}\mathrm{cos}(2x) + \frac{3}{4}x^{2}\mathrm{cos}(2x) + \frac{3}{4}x\mathrm{cos}(2x) - \frac{3}{8}\mathrm{sin}(2x) + C \end{aligned}\]38. \(\int x^{2}\sqrt[3]{1 + x^{3}} \ dx\)
\[\begin{aligned} \int x^{2}\sqrt[3]{1 + x^{3}} \ dx &= \frac{1}{3}\int 3x^{2}\sqrt[3]{1 + x^{3}} \ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} u &= 1 + x^{3}\\ du &= 3x^{2}\ dx\\ dx &= \frac{1}{3x^{2}}\ du \end{aligned}\] \[\begin{aligned} \frac{1}{3}\int 3x^{2}\sqrt[3]{1 + x^{3}} \ dx &= \frac{1}{3}\int 3x^{2}\sqrt[3]{u}\times \frac{1}{3x^{2}}\ du\\ &= \frac{1}{3}\int \sqrt[3]{u} \ du\\ &= \frac{1}{3}\times\frac{3}{4}u^{\frac{4}{3}} + C\\ &= \frac{1}{4}\sqrt[3]{(1 + x^{3})^{4}} + C\\ &= \frac{1}{4}(1 + x^{3})\sqrt[3]{1 + x^{3}} + C \end{aligned}\]39. \(\int \frac{1}{(x^{2} + 4)^{2}}\ dx\)
\[\begin{aligned} \int \frac{1}{(x^{2} + 4)^{2}}\ dx \end{aligned}\]Trig Substitution:
\[\begin{aligned} x &= 2 \mathrm{tan}(\theta)\\ dx &= 2 \mathrm{sec}^{2}(\theta)\ d\theta \end{aligned}\] \[\begin{aligned} \int \frac{1}{(x^{2} + 4)^{2}}\ dx &= \int \frac{1}{(2^{2}\mathrm{tan}^{2}(\theta) + 4)^{2}} \times 2 \mathrm{sec}^{2}(\theta)\ d\theta\\ &= \int \frac{1}{(4\mathrm{tan}^{2}(\theta) + 4)^{2}} \times 2 \mathrm{sec}^{2}(\theta)\ d\theta\\ &= \int \frac{1}{(4(\mathrm{tan}^{2}(\theta) + 1))^{2}} \times 2 \mathrm{sec}^{2}(\theta)\ d\theta\\ &= \int \frac{1}{(4\mathrm{sec}^{2}(\theta))^{2}}\ dx \times 2 \mathrm{sec}^{2}(\theta)\ d\theta\\ &= \int \frac{1}{16\mathrm{sec}^{4}(\theta)}\ dx \times 2 \mathrm{sec}^{2}(\theta)\ d\theta\\ &= \frac{1}{8}\int \mathrm{cos}^{2}\theta \ d\theta\\ &= \frac{1}{8}\times\frac{1}{2}\int 1 + \mathrm{cos}(2\theta)\ d\theta\\ &= \frac{1}{16}\Big(\theta + \frac{1}{2}\mathrm{sin}(2\theta)\Big)\\ &= \frac{1}{16}\Big(\theta + \frac{1}{2}2\mathrm{sin}(\theta)\mathrm{cos}(\theta)\Big)\\ &= \frac{1}{16}\theta + \frac{1}{16}\mathrm{sin}(\theta)\mathrm{cos}(\theta)[5pt] \end{aligned}\] \[\begin{aligned} \mathrm{tan}(\theta) &= \frac{x}{2}\\ \mathrm{sin}(\theta) &= \frac{x}{\sqrt{x^{2} + 4}}\\ \mathrm{cos}(\theta) &= \frac{2}{\sqrt{x^{2} + 4}}\\ \theta &= \mathrm{tan}^{-1}\Big(\frac{x}{2}\Big)[5pt] \end{aligned}\] \[\begin{aligned} \frac{1}{16}\theta + \frac{1}{16}\mathrm{sin}(\theta)\mathrm{cos}(\theta) &= \frac{1}{16}\mathrm{tan}^{-1}\Big(\frac{x}{2}\Big) + \frac{1}{16}\frac{x}{\sqrt{x^{2} + 4}}\frac{2}{\sqrt{x^{2} + 4}}\\ &= \frac{1}{16}\mathrm{tan}^{-1}\Big(\frac{x}{2}\Big) + \frac{1}{8}\frac{x}{x^{2} + 4} \end{aligned}\]40. \(\int_{1}^{2} \sqrt{x^{2} - 1} \ dx\)
\[\begin{aligned} \int_{1}^{2} \sqrt{x^{2} - 1} \ dx \end{aligned}\]Let:
\[\begin{aligned} y^{2} &= x^{2} - 1\\ \frac{x^{2}}{1} - \frac{y^{2}}{1} &= 1 \end{aligned}\]This is a hyperbola equation. From the hyperbolic function wiki link, we can see the following property (image taken from same wiki link):
Taking the point:
\[\begin{aligned} x &= 2\\ &= \mathrm{cosh}(t)\\ y &= \sqrt{2^{2} - 1}\\ &= \sqrt{3}\\ &= \mathrm{sinh}(t) \end{aligned}\] \[\begin{aligned} t &= \mathrm{cosh}^{-1}(2)\\ t &= \mathrm{sinh}^{-1}(\sqrt{3}) \end{aligned}\] \[\begin{aligned} \int_{1}^{2} \sqrt{x^{2} - 1} \ dx &= \frac{1}{2} \times 2 \times\sqrt{3} - \frac{t}{2}\\ &= \sqrt{3} - \frac{\mathrm{cosh}^{-1}(2)}{2}\\ &= \sqrt{3} - \frac{\mathrm{sinh}^{-1}(\sqrt{3})}{2} \end{aligned}\]