5.5 The Substitution Rule/21: Difference between revisions

Latest revision as of 23:27, 13 September 2022

$\int {\frac {\cos {({\sqrt {t}})}}{\sqrt {t}}}\;dt$

${\begin{aligned}u&={\sqrt {t}}\\[2ex]du&=({\frac {1}{2}}\ {\frac {1}{\sqrt {t}}})\;dt\\[2ex]2du&={\frac {1}{\sqrt {t}}}\;dt\end{aligned}}$

${\begin{aligned}\int {\frac {1}{\sqrt {t}}}\cos {({\sqrt {t}})}dt&=2\int \cos {u}\;du\\[2ex]&=2\sin {u}+c\\[2ex]&=2\sin({\sqrt {t}})+c\\[2ex]\end{aligned}}$

@@ Line 1: / Line 1: @@
 <math>
-\int \frac{\cos{(\sqrt{t})}}{\sqrt{t}}dt
+\int \frac{\cos{(\sqrt{t})}}{\sqrt{t}}\;dt
-\int\sqrt{u}du \\[2ex]
 </math>
@@ Line 9: / Line 7: @@
 \begin{align}
-& \int\frac{\left(\ln(x)\right)^2}{x}dx \ = \ \int u^2du \\[2ex]
+u &= \sqrt{t} \\[2ex]
-& = \ \frac{u^{2+1}}{2+1}du \ = \ \frac{1}{3}u^3+C \\[2ex]
+du &= (\frac{1}{2}\ \frac{1}{\sqrt{t}})\;dt \\[2ex]
+du &= \frac{1}{\sqrt{t}}\;dt
+\end{align}
+</math>
-& u=\ln(x) \\
-& du=\frac{1}{x}dx \\
+<math>
+\begin{align}
-& = \ \frac{1}{3}(\ln(x))^3+C
+\int \frac{1}{\sqrt{t}}\cos{(\sqrt{t})} dt &= 2\int \cos {u}\;du \\[2ex]
+&= 2 \sin{u}+c \\[2ex]
+&= 2 \sin(\sqrt{t}) + c \\[2ex]
 \end{align}
 </math>

5.5 The Substitution Rule/21: Difference between revisions

Latest revision as of 23:27, 13 September 2022

Navigation menu

Search