5.5 The Substitution Rule/37: Difference between revisions

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u &= \sin(x) \\[2ex]
u &= \sin(x) \\[2ex]
du &= (\cos(x))dx \\[2ex]
du &= \cos(x)\;dx \\[2ex]
\frac{1}{3}du &= (a+bx^2)dx \\[2ex]


\end{align}
\end{align}
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\begin{align}
\begin{align}


\int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx &= \int \frac{1}{\sqrt{3ax+bx^3}}(a+bx^2)\;dx = \int \frac{1}{\sqrt{3ax+bx^3}}(a+bx^2\;dx)\ \\[2ex]
\int \frac{\cos(x)}{\sin(x)}dx &= \int \frac{1}{\sin(x)}\cos(x)\;dx = \int \frac{1}{\sin(x)}(\cos(x)\;dx)  \\[2ex]




&= \frac{1}{3}\int \frac{1}{\sqrt{u}}(du) = \frac{1}{3}\int u^{-1/2} du \\[2ex]
&= \int \frac{1}{{u}}(du) \\[2ex]
&= \frac{1}{3}\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C \\[2ex]
&= \left| \ln(u) \right| + C \\[2ex]
&= \frac{2}{3}(3ax+bx^3)^{1/2} + C \\[2ex]
&= \left| \ln(\sin(x)) \right| + C \\[2ex]
&= \frac{2}{3}{\sqrt{3ax+bx^3}} + C
 


\end{align}
\end{align}
</math>
</math>

Latest revision as of 19:18, 20 September 2022