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March 11th, 2017, 06:19 PM   #1
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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
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March 11th, 2017, 06:57 PM   #2
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(2). Completing the square, -5 - 3x² + 12x = 7 - 3(x - 2)², so use x = √(7/3)sin(t) + 2.
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March 11th, 2017, 07:50 PM   #3
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Quote:
Originally Posted by jackhammer View Post
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.
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