How do I evaluate the integral #intsqrt(54+9x^2)dx#?

Question

William Abdalla · Accepted Answer

#intsqrt(9x^2+54)dx=(3sqrt6)/2xsqrt(x^2/6+1)+9ln(|(sqrt6x)/6+sqrt(x^2/6+1)|)+C#, #C in RR#

#I=intsqrt(9x^2+54)dx#

#=intsqrt(9(x^2+6)dx#

#=3intsqrt(x^2+6)dx#

Let #x=sqrt6tan(theta)#

#dx=sqrt6sec(theta)^2d theta#

So:

#I=3intsqrt((sqrt6tan(theta))^2+6)*sqrt6sec(theta)^2d theta#

#=3sqrt6intsqrt(6(tan(theta)^2+1))sec(theta)^2d theta#

Because #tan(theta)^2+1=sec(theta)^2#

#I=18intsec(theta)^3d theta=18intsec(theta)*sec(theta)^2d theta#

Using Integration by parts :

#intf'(x)g(x)dx=f(x)g(x)-intf(x)g'(x)dx#

Here: #f'(x)=sec(theta)^2<=>f(x)=tan(theta)#

#g(x)=sec(theta)<=>g'(x)=tan(theta)sec(theta)# So:

#I=18tan(theta)sec(theta)-18inttan(theta)^2sec(theta)d theta#

#=18tan(theta)sec(theta)-18int(sin(theta)^2)/(cos(theta)^3)d theta#

#=18tan(theta)sec(theta)-18int(1-cos(theta)^2)/(cos(theta)^3)d theta#

#=18tan(theta)sec(theta)-18int1/(cos(theta)^3)d theta+18int1/cos(theta)d theta#

But #I=18intsec(theta)^3d theta#

So: #I=18tan(theta)sec(theta)-I+18##intsec(theta)d theta#

#2I=18tan(theta)sec(theta)+18ln(|tan(theta)+sec(theta)|)#

#I=9tan(theta)sec(theta)+9ln(|tan(theta)+sec(theta)|)#

Finally, because #theta=tan^(-1)((sqrt6x)/6)# and that #sec(tan^(-1)(u))=sqrt(u^2+1)#,

#I=(3sqrt6)/2xsqrt(x^2/6+1)+9ln(|(sqrt6x)/6+sqrt(x^2/6+1)|)+C#, #C in RR#

\0/ Here's our answer !

Note : click on each blue terms to see more explanations which were secondary but still important.

Ezekiel Alexander · Answer

#int(54+9x^2)dx=3/2{xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|}+C#

Here ,

#color(blue)(intsqrt(54+9x^2)dx=3intsqrt(6+x^2)dx=3I............(A)#

Let ,

#color(red)(I=intsqrt(6+x^2)dx.........................to(B)#

#I=intsqrt(6+x^2)*1dx#

Using Integration by parts:

#I=sqrt(6+x^2)int1dx-int(d/(dx)(sqrt(6+x^2))int1dx)dx#

#I=sqrt(6+x^2)xx x-int(2x)/(2sqrt(6+x^2)) xx xdx#

#I=sqrt(6+x^2)xx x-int(x^2)/(sqrt(6+x^2)) dx#

#I=sqrt(6+x^2)xx x-int(6+x^2-6)/(sqrt(6+x^2)) dx#

#I=sqrt(6+x^2)xx x-int(6+x^2)/sqrt(6+x^2)dx+int6/(sqrt(6+x^2)) dx#

#I=xsqrt(6+x^2)-intsqrt(6+x^2)dx+6color(green)(int1/(sqrt((sqrt6)^2+x^2)) dx)#

#I=xsqrt(6+x^2)-intsqrt(6+x^2)dx+6color(green)(ln|x+sqrt(6+x^2)|)+c#

#I=xsqrt(6+x^2)-color(red)(I)+6ln|x+sqrt(6+x^2)|+c, color(red)(....touse(B)#

#I+I=xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|+c#

#:.2I=xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|+c#

#:.I=1/2{xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|}+c/2#

From #color(blue)((B)# we have ,

#int(54+9x^2)dx=3/2{xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|}+(3c)/2#

#int(54+9x^2)dx=3/2{xsqrt(6+x^2)+6ln|x+sqrt(6+x^2)|}+C#

Note :

#color(green)(int1/sqrt(X^2+a^2)dX=ln|X+sqrt(X^2+a^2)|+c#

Paisley Johnson · Answer

undefined To evaluate the integral ∫√(54 + 9x^2) dx, you can use trigonometric substitution. First, rewrite the expression under the square root as 9(6 + x^2/2^2). Then, let x = 2tanθ and dx = 2sec^2θdθ. Substitute these into the integral and simplify to integrate in terms of θ. Finally, reintroduce x into the result and simplify. The final answer will be in terms of x.