How do you integrate #int sqrt(x^2-25) dx# using trigonometric substitution?

Answer 1

#color(green)[int sqrt(x^2-25)*dx=-(25*ln(abs(sqrt(x^2-25)+x))-x*sqrt(x^2-25))/2]#

The integral is #int sqrt(x^2-25)*dx#

Let suppose:

#x=5sectheta#
#dx=5*sectheta*tantheta*d(theta)#
#sqrt(x^2-25)=sqrt[25(sec^2theta-1)]=5tantheta#

the integral become after suppose:

#int sqrt(x^2-25)*dx=int5tantheta*5sectheta*tantheta*d(theta)#
#25intsectheta*tan^2theta*d(theta)=25intsectheta*(sec^2theta-1)*d(theta)#
#25intsec^3theta-sectheta*d(theta)=25intsec^3theta-25intsectheta*d(theta)#
Firstly let solve #color(red)[intsec^3theta*d(theta)# by using integral y parts:
#color(red)(I=intsec^3(theta)d(theta)...to(1)#
#"Using "color(blue)"Integration by Parts"#
#int(u*v)dx=uintvdx-int(u'intvdx)d(theta)#
Let #u=sec(theta) and v=sec^2(theta)#
#=>u'=sec(theta)tan(theta) and intvd(theta)=tan(theta)#
#I=sec(theta) xxtan(theta)-intsec(theta)tantheta xxtan(theta) d(theta)#
#=secthetatan(theta)-intsec(theta)tan^2(theta)dx+c#
#=sec(theta)tan(theta)-intsec(theta)(sec^2(theta)-1)d(theta)+c#
#I=sec(theta)tan(theta)-color(red)(intsec^3(theta)dx)+intsec(theta)d(theta)+c#
#I=sec(theta)tan(theta)-color(red)(I)+intsec(theta)d(theta)+c...to#from1
#I+I=sec(theta)tan(theta)+intsec(theta)d(theta)+c#
#2I=sec(theta)tan(theta)+ln|sec(theta)+tan(theta)|+C#
#I=1/2sec(theta)tan(theta)+1/2ln|sec(theta)+tan(theta)|+C#
secondly let find #color(blue)[intsectheta*d(theta)#
#intsectheta*d(theta)=intsectheta*[sectheta+tantheta]/[sectheta+tantheta]*d(theta)#
#int(sec^2theta+sectheta*tantheta)/(sectheta+tantheta)*d(theta)=ln|sectheta+tantheta|+c#
#color(green)[25intsec^3theta-25intsectheta*d(theta)]=#
#25[1/2sec(theta)tan(theta)+1/2ln|sec(theta)+tan(theta)|]-25[ln|sectheta+tantheta|]+c#
#=(x*sqrt(x^2-25))/2-(25*ln(abs(sqrt(x^2-25)+x)))/2#

After simplified it we will get:

#=-(25*ln(abs(sqrt(x^2-25)+x))-x*sqrt(x^2-25))/2#
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Answer 2

To integrate ∫√(x^2 - 25) dx using trigonometric substitution, we can use x = 5 sec(θ). Then dx = 5 sec(θ) tan(θ) dθ. Substituting these expressions into the integral and simplifying yields ∫(5 sec(θ))^2 tan(θ) dθ. Simplify further to obtain ∫25 sec^2(θ) tan(θ) dθ. Using the identity sec^2(θ) = tan(θ) + 1, the integral becomes ∫25(tan(θ) + 1) tan(θ) dθ. This simplifies to ∫(25tan^2(θ) + 25) dθ. Integrate term by term to get 25∫tan^2(θ) dθ + 25∫dθ. Using the trigonometric identity tan^2(θ) = sec^2(θ) - 1, the integral becomes 25∫(sec^2(θ) - 1) dθ + 25∫dθ. Integrate each term to get 25(tan(θ) - θ) + 25θ + C. Substitute back x = 5 sec(θ) to obtain the final result: 25(tan^(-1)(x/5) - x/5) + 25 tan^(-1)(x/5) + C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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