Find the Integral using Trigonometric Substitution
- Thread starter Glazer
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For type-setting clarity, I believe the above indicates the following:
. . . . .\(\displaystyle \int\, \dfrac{x^2\, +\, 1}{(x^2\, -\, 2x\, +\, 2)^2}\, dx\)
A good start would be to follow the first few steps displayed in the worked examples in your book and in your class notes (and in online resources, such as in this listing of lessons). Note: I suspect that you actually mean to refer to inverse-trig substitutions, such as are explained and illustrated here...?
What techniques and tactics do you notice them using in the various relevant worked examples? Does "completing the square" ring any bells?
I googled the problem earlier and found someone used these steps:
I understand everything above. (u=x-1 and x=1+u)
Using the substitution u=tan(y) gives
I dont understand why he substitutes u with tan(y). I understand that when given the expression [FONT=Georgia, Times New Roman, Times, serif]sqrt(a2+x2), x=a*tan(y). Im just confused because there is no square root in this.
[/FONT]Also, for future reference, what do you use to write integrals? I just copied and pasted the above ones.
∫x2+1(x2−2x+2)2dx=∫x2+1((x−1)2+1)2dx=∫(1+u)2+1(u2+1)2du.
Using the substitution u=tan(y) gives
∫tan(y)2+2tan(y)+2(1+tan(y)2)2
dy
dy
[/FONT]Also, for future reference, what do you use to write integrals? I just copied and pasted the above ones.
skeeter
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\(\displaystyle \displaystyle \int \frac{x^2+1}{(x^2-2x+2)^2} \, dx\)
\(\displaystyle \displaystyle \int \frac{x^2+1}{[(x^2-2x+1)+1]^2} \, dx\)
\(\displaystyle \displaystyle \int \frac{x^2+1}{[(x-1)^2+1]^2} \, dx\)
\(\displaystyle \displaystyle x-1 = \tan{t} \implies x = \tan{t} + 1 \implies dx = \sec^2{t} \, dt\)
\(\displaystyle \displaystyle \int \frac{(\tan{t}+1)^2+1}{(\tan^2{t}+1)^2} \cdot \sec^2{t} \, dt\)
\(\displaystyle \displaystyle \int \frac{\tan^2{t}+2\tan{t}+2}{(\sec^2{t})^2} \cdot \sec^2{t} \, dt\)
\(\displaystyle \displaystyle \int \frac{\tan^2{t}+2\tan{t}+2}{\sec^2{t}} \, dt\)
\(\displaystyle \displaystyle \int \sin^2{t} + 2\sin{t}\cos{t} + 2\cos^2{t} \, dt\)
\(\displaystyle \displaystyle \int \sin(2t) + \cos^2{t} + 1 \, dt\)
\(\displaystyle \displaystyle \int \sin(2t) + \frac{1+\cos(2t)}{2} + 1 \, dt\)
\(\displaystyle \displaystyle \int \sin(2t) + \frac{\cos(2t)}{2} + \frac{3}{2} \, dt\)
\(\displaystyle \displaystyle -\frac{\cos(2t)}{2} + \frac{\sin(2t)}{4} + \frac{3t}{2} + C\)
\(\displaystyle \displaystyle \tan{t} = x-1 \implies \sin{t} = \frac{x-1}{\sqrt{x^2-2x+2}} \implies \cos{t} = \frac{1}{\sqrt{x^2-2x+2}}\)
\(\displaystyle \displaystyle \cos(2t) = 2\cos^2{t}-1 = \frac{2}{x^2-2x+2} - 1\)
\(\displaystyle \displaystyle \sin(2t) = 2\sin{t}\cos{t} = \frac{2(x-1)}{x^2-2x+2}\)
\(\displaystyle \displaystyle t = \arctan(x-1)\)
so, after a little more grunt work algebra ...
\(\displaystyle \displaystyle \int \frac{x^2+1}{(x^2-2x+2)^2} \, dx = \frac{x-3}{2(x^2-2x+2)}+\frac{3}{2}\arctan(x-1)+C\)
\(\displaystyle \displaystyle \int \frac{x^2+1}{[(x^2-2x+1)+1]^2} \, dx\)
\(\displaystyle \displaystyle \int \frac{x^2+1}{[(x-1)^2+1]^2} \, dx\)
\(\displaystyle \displaystyle x-1 = \tan{t} \implies x = \tan{t} + 1 \implies dx = \sec^2{t} \, dt\)
\(\displaystyle \displaystyle \int \frac{(\tan{t}+1)^2+1}{(\tan^2{t}+1)^2} \cdot \sec^2{t} \, dt\)
\(\displaystyle \displaystyle \int \frac{\tan^2{t}+2\tan{t}+2}{(\sec^2{t})^2} \cdot \sec^2{t} \, dt\)
\(\displaystyle \displaystyle \int \frac{\tan^2{t}+2\tan{t}+2}{\sec^2{t}} \, dt\)
\(\displaystyle \displaystyle \int \sin^2{t} + 2\sin{t}\cos{t} + 2\cos^2{t} \, dt\)
\(\displaystyle \displaystyle \int \sin(2t) + \cos^2{t} + 1 \, dt\)
\(\displaystyle \displaystyle \int \sin(2t) + \frac{1+\cos(2t)}{2} + 1 \, dt\)
\(\displaystyle \displaystyle \int \sin(2t) + \frac{\cos(2t)}{2} + \frac{3}{2} \, dt\)
\(\displaystyle \displaystyle -\frac{\cos(2t)}{2} + \frac{\sin(2t)}{4} + \frac{3t}{2} + C\)
\(\displaystyle \displaystyle \tan{t} = x-1 \implies \sin{t} = \frac{x-1}{\sqrt{x^2-2x+2}} \implies \cos{t} = \frac{1}{\sqrt{x^2-2x+2}}\)
\(\displaystyle \displaystyle \cos(2t) = 2\cos^2{t}-1 = \frac{2}{x^2-2x+2} - 1\)
\(\displaystyle \displaystyle \sin(2t) = 2\sin{t}\cos{t} = \frac{2(x-1)}{x^2-2x+2}\)
\(\displaystyle \displaystyle t = \arctan(x-1)\)
so, after a little more grunt work algebra ...
\(\displaystyle \displaystyle \int \frac{x^2+1}{(x^2-2x+2)^2} \, dx = \frac{x-3}{2(x^2-2x+2)}+\frac{3}{2}\arctan(x-1)+C\)