5.5 The Substitution Rule/35: Difference between revisions

Latest revision as of 05:31, 5 September 2022

${\begin{aligned}\ \int {\frac {sin2x}{1+cos^{2}x}}dx\end{aligned}}$

${\begin{aligned}let\;u&=1+cos^{2}x\\[2ex]du&=2cosx\;\cdot (-sinx)\;dx\\[2ex]-du&=2sin(x)cos(x)\;dx&\\[2ex]-du&=sin(2x)dx&\end{aligned}}$

${\begin{aligned}\ \int {\frac {sin2x}{1+cos^{2}x}}dx\;=\;-\ \int {\frac {du}{u}}\;=\;-ln(1+u)+c\;=\;-|1+cos^{2}x|+c\end{aligned}}$

@@ Line 4: / Line 4: @@
 <math>\begin{align}
-let \; u &=1 + cos^2x \\[2ex]
+let \; u &=1 + cos^2x  \\[2ex]
-du &= 2cosx \;\cdot (-sinx) \\[2ex]
+du &= 2cosx \;\cdot (-sinx)\;dx \\[2ex]
 -du & = 2sin(x)cos(x)\;dx & \\[2ex]
 -du & = sin(2x)dx &
@@ Line 12: / Line 12: @@
-<math>\begin{align} \ \int \frac{sin 2 x}{1 + cos^2x} dx = -\ \int \frac {du}{u} = \ \int -ln( 1 + u)= \ \int| 1 + cos^2x| + c
+<math>\begin{align} \ \int \frac{sin 2 x}{1 + cos^2x} dx\;=\; -\ \int \frac {du}{u}\;=\;-ln( 1 + u) + c\;=\;-| 1 + cos^2x| + c
 \end{align}
 </math>