My Math Forum Integrals and trigonometric substitution

 Calculus Calculus Math Forum

 March 11th, 2017, 05:19 PM #1 Newbie   Joined: Mar 2017 From: canada Posts: 2 Thanks: 0 Integrals and trigonometric substitution Hi guys, Can you help me with the following: Find the appropriate trigonometric substitution of the form x=f(t) to simplify the integral. 1)Integral x*Sqrt(3x^2+12x+6) dx x=? 2)Integral x/sqrt(-5-3x^2+12x) dx x=? Thanks
 March 11th, 2017, 05:57 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 (2). Completing the square, -5 - 3x² + 12x = 7 - 3(x - 2)², so use x = √(7/3)sin(t) + 2.
March 11th, 2017, 06:50 PM   #3
Member

Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

Quote:
 Originally Posted by jackhammer Hi guys, Can you help me with the following: Find the appropriate trigonometric substitution of the form x=f(t) to simplify the integral. 1)Integral x*Sqrt(3x^2+12x+6) dx x=? 2)Integral x/sqrt(-5-3x^2+12x) dx x=? Thanks
\displaystyle \begin{align*} \int{ x\,\sqrt{3\,x^2 + 12\,x + 6}\,\mathrm{d}x} &= \sqrt{3}\,\int{ x\,\sqrt{ x^2 + 4\,x + 2 }\,\mathrm{d}x } \\ &= \frac{\sqrt{3}}{2}\,\int{ 2\,x\,\sqrt{ x^2 + 4\,x + 2 }\,\mathrm{d}x } \\ &= \frac{\sqrt{3}}{2}\,\int{ \left[ \left( 2\,x + 4 \right) \sqrt{ x^2 + 4\,x + 2 } - 4\,\sqrt{ x^2 + 4\,x + 2 } \right] \,\mathrm{d}x } \\ &= \frac{\sqrt{3}}{2}\,\int{ \left( 2\,x + 4 \right) \sqrt{x^2 + 4\,x + 2} \,\mathrm{d}x } - 2\,\sqrt{3}\,\int{ \sqrt{x^2 + 4\,x + 2}\,\mathrm{d}x } \\ &= \frac{\sqrt{3}}{2}\,\int{ \left( 2\,x + 4 \right) \sqrt{x^2 + 4\,x + 2}\,\mathrm{d}x } - 2\,\sqrt{3}\,\int{ \sqrt{\left( x + 2 \right) ^2 - 2}\,\mathrm{d}x } \end{align*}

In the first integral let \displaystyle \begin{align*} u = x^2 + 4\,x + 2 \implies \mathrm{d}u = \left( 2\,x + 4 \right) \,\mathrm{d}x \end{align*} and in the second integral let \displaystyle \begin{align*} x + 2 = 2 \cosh{(t)} \implies \mathrm{d}x = 2 \sinh{(t)}\,\mathrm{d}t \end{align*} and we get

\displaystyle \begin{align*} \frac{\sqrt{3}}{2}\,\int{ \left( 2\,x + 4 \right) \sqrt{x^2 + 4\,x + 2}\,\mathrm{d}x } - 2\,\sqrt{3}\,\int{ \sqrt{\left( x + 2 \right) ^2 - 2}\,\mathrm{d}x } &= \frac{\sqrt{3}}{2}\,\int{ \sqrt{u}\,\mathrm{d}u } - 2\,\sqrt{3}\,\int{\sqrt{\left[ 2\cosh{(t)} \right]^2 - 2^2}\,\left[ 2\sinh{(t)} \right] \,\mathrm{d}t} \\ &= \frac{\sqrt{3}}{2}\,\int{ u^{\frac{1}{2}}\,\mathrm{d}u } - 4\,\sqrt{3}\,\int{ 2\sinh^2{(t)}\,\mathrm{d}t } \\ &= \frac{\sqrt{3}}{2}\,\left( \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C_1 \right) - 4\,\sqrt{3}\,\int{ \left[ \cosh{(2\,t)} - 1 \right] \,\mathrm{d}t } \\ &= \frac{\sqrt{3}}{3}\,u^{\frac{3}{2}} + \frac{\sqrt{3}}{2}\,C_1 - 4\,\sqrt{3}\,\left[ \frac{1}{2}\,\sinh{(2\,t)} - t + C_2 \right] \\ &= \frac{\sqrt{3}}{2}\,\left( x^2 + 4\,x + 2 \right) ^{\frac{3}{2}} + \frac{\sqrt{3}}{2} \,C_1 - 2\,\sqrt{3}\,\sinh{(2\,t)} + 4\,\sqrt{3}\,t - 4\,\sqrt{3}\,C_2 \\ &= \frac{\sqrt{3}}{2}\,\left( x^2 + 4\,x + 2 \right) ^{\frac{3}{2}} - \sqrt{3}\,\sinh{(t)}\cosh{(t)} - 4\,\sqrt{3}\,t + C \textrm{ where } t = \frac{\sqrt{3}}{2}\,C_1 - 4\,\sqrt{3}\,C_2 \\ &= \frac{\sqrt{3}}{2}\,\left( x^2 + 4\,x + 2 \right) ^{\frac{3}{2}} - \sqrt{3}\,\sqrt{ \left( \frac{x + 2}{2} \right) ^2 - 1 }\,\left( \frac{x + 2}{2} \right) - \textrm{arcosh}\,\left( \frac{x + 2}{2} \right) + C \end{align*}

and now simplify further if you wish.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post devinperry Calculus 4 April 12th, 2014 12:46 PM cheyb93 Calculus 3 January 19th, 2013 06:27 PM Jet1045 Calculus 3 March 8th, 2012 11:22 AM izseekzu Calculus 1 February 16th, 2010 08:09 PM mmmboh Calculus 8 October 26th, 2008 07:46 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top