- #1
KDeep
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Homework Statement
integral ofdx/((9-(x^2))^(3/2)) A = 0, B = 3/2
Homework Equations
Trigonometry Substitutions3. The Attempt at a Solution : I am stuck with this question. So far, I got
(1/9)integral of (1/cos^2(θ)) dθ
KDeep said:That's my problem. My trigonometry is very limited. I do not know of a Anti-derivative of sec^2θ
KDeep said:the derivative of tan(∅) is sec^2(∅). I get 1/9(tan((sin(∅))) | , Could someone solve the entire problem for me ? and reply with the steps to get there.
This would be a lot of help!
KDeep said:A = 0 , B = 3/2 ∫ dx/((9-x2)3/2)
First I substituted , x = 3sin∅ , dx = 3cos∅d∅
∫ 3cos∅d∅/(32-(3sin∅)2)3/2
Then I factored.
∫ 3cos∅d∅/(32(1-sin2∅))3/2
Then I used the trig identity to convert 1-sin2∅ to cos2θ
∫ 3cos∅d∅/(32cos2∅)3/2
Then I simplified
(1/9) ∫ (1/cos2θ) d∅
Then I anti-derived that into..
(1/9) tanθ | [ I do not know what to do after this, Since it is respect to ∅]
To conclude, I cannot use A and B, Since those are respect to x. I am stuck on this part.
ammus said:Can you tell me how to crack the following problems?:
1) y=2x(dy/dx)-y(dy/dx)2
2) (d2y/dx2)+y=2 cos2(x)+sin(3x)
KDeep said:Ok that's what I was going for.
So tan(sin-1(x/3))
(sin-1((3/2)/3)) is ∏/6 right?1/9 (Tan (∏/6) - Tan(0)) ? Am I on the right track?
KDeep said:Yeah thanks for the tip.
So the answer is 1/(9√3)? Am I correct?
Trig Substitution is a technique used in integration to solve integrals that involve trigonometric functions. It involves substituting a trigonometric function or expression for a variable in the integral.
Trig Substitution is typically used when the integral involves expressions with sqrt(a^2 - x^2), sqrt(x^2 + a^2), or sqrt(x^2 - a^2). It is also useful when the integrand contains a quadratic expression in terms of sine or cosine.
The choice of trigonometric substitution depends on the form of the integral. For sqrt(a^2 - x^2), use x = a sin(theta). For sqrt(x^2 + a^2), use x = a tan(theta). And for sqrt(x^2 - a^2), use x = a sec(theta). It is important to choose the substitution that will cancel out the square root term and simplify the integral.
Yes, there are other techniques such as Integration by Parts, Partial Fractions, and Trigonometric Identities. However, Trig Substitution is specifically used for integrals involving trigonometric functions and can be a more efficient method in those cases.
Yes, Trig Substitution can be used for both indefinite and definite integrals. When using Trig Substitution for a definite integral, be sure to adjust the limits of integration to match the new variable, and then evaluate the integral as usual.