Trig substitution Definition and 117 Threads

\int \frac{x^2}{\sqrt{9-x^2}} [SIZE="4"] find the integral using trig sub x= 3 \sin {\phi} replace 3sin\phi into x and solve. I got to \int \frac{9-9 \cos{\phi}}{3 \cos{\phi}} then what should I do?

Homework Statement Evaluate ∫ x √ 4 + x2 dx by using the trigonometric substitution x = 2tanθ I am starting on the right track by subbing x=2tanθ into x like this: =∫ 2tanθ √ 4 + 2tanθ(2) then, do I just integrate that for the correct answer?

\int_{-2} ^2 \frac{dx}{4+x^2} I use the trig substitution and get everything done but for some reason I can't get the answer, here's all my working: x = 2 \tan\theta dx = 2 \sec^2\theta 4+x^2=4(1+\tan\theta)=4\sec^2\theta \int \frac{2\sec^2\theta d\theta}{4\sec^2\theta} \int...

The question is to evaluate the integral in the attachment. Using trig substition, I've reduced it to ∫ (tanz)^2 where z will be found using the triangle. I just need to integrate tangent squared which I can't seem to figure how to do. I tried using the trig identity (secx)^2 - 1 but I don't...

Homework Statement \int\frac{dx}{\sqrt{x^2 + 16}}Homework Equations x = 4\tan\theta dx = 4\sec^2\theta \ d\thetaThe Attempt at a Solution \int\frac{4\sec^2\theta}{\sqrt{16\tan^2\theta + 16}}\ d\theta \int\frac{4\sec^2\theta}{\sqrt{16(\tan^2\theta + 1)}}\ d\theta...

Homework Statement Evaluate the following integrals or state that they diverge. Use proper notation. Integral from 0 to 2 of (x+1)/Square root(4-x^2) Homework Equations The Attempt at a Solution I just substituted x = 2sin(theta) thus dx = 2cos(theta) I got to the...

Homework Statement Homework Equations The Attempt at a Solution I'm not asking for someone to do the question for me but I was just wondering what I'm supposed to sub in. Do I put in as if it was (x^2-9)^(1/2) or do I have to do something differently if there is a constant in front...

The bit of the problem that I'm working on: 6\int\frac{dx}{x^2-x+1} My work: =6\int\frac{dx}{(x^2-x+\frac{1}{4})+1-\frac{3}{4}} =6\int\frac{dx}{(x-\frac{1}{2})^2+\sqrt{\frac{3}{4}}^2} let x-\frac{1}{2}=\sqrt{\frac{3}{4}}\tan\theta so dx=\sqrt{\frac{3}{4}}\sec^2\theta d\theta...

How would I go about solving this: \int cos^2x tan^3xdx..all i did so far .. \int cos^2x tanx(tan^2x) [[ \frac{1}{cosx}^2 -1] = tan^2x so... \int cos^2x[[ \frac{1}{cosx}]^2-1]tanx is this right so far...now what?

I am working on a Differential Equation problem and I am stuck on these two integrals: http://forums.cramster.com/Answer-Board/Image/cramster-equation-20064101738436328028752372062504976.gif and...

I am not too good with trig identities. I can't seem to figure out how to simplify these trig intergrals. I know I can use a triangle to turn the second problem into a trig integral, but once I have the trig integral, I am lost. Any help would be greatly appriciated.:redface...

On this physics problem i need to do a double integral (dx,dy) of 1/sqrt(x^2 + y^2 +z^2). Which looks easy enough at first, until I reallized (after many hours) I cannot figure out how to integrate it. I am sure at this point there is some trig substitution (learned too long ago...), but I am...

When is th ebest time to use it and what are some good rules of thumb for it?

im hoping i worked this out right; its long: \int x(81-x^2)^{5/2}dx the integral contains a^2-x^2, so i set x=asin\theta. that would make x=9sin\theta and dx=9cos\theta d\theta: \int 9sin\theta(81-81sin^2\theta)^{5/2}9cos\theta d\theta = \int 9sin\theta[81(1-sin^2\theta)]^{5/2}9cos\theta...

Use trig substitution to find \int_{0}^{5} \frac{dt}{25 + x^2}dt I can solve it to here \int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{(25 + tan^2\theta)^2} and from this point i can factor the denominator into {625(1+ \tan^2\theta)}^2 which becomes 625\sec^4\theta now i have the...

can someone help me find a appropriate trig sub for this problem: \int\frac{x}{sqrt(-29-4x^2-24x)} took out sqrt(4)... sqrt(4)*sqrt(-29/4-x^2-6x) (i also changed all the negative signs to positive) complete the square... sqrt(4)*sqrt((x+3)^2-7/4) so my trig sub should be...

How would you go about solving \int \frac{\sqrt{1-x^2}}{x^2} ? I have tried a few things... drawing out triangles... etc but can't seem to get it... I am kind of behind in math because I was gone for awhile because of being sick and presentations.