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Integral
- Thread starter steve.b
- Start date
Hello, steve.b!
The second one uses Partial Fractions . . .
\(\displaystyle \text{We have: }\;\frac{x^2+x+1}{(x+1)(x^2+2x+2)} \;=\;\frac{A}{x+1} + \frac{Bx}{x^2+2c+2} + \frac{C}{x^2+2x+2}\)
. . . . . . . . . . . . . . \(\displaystyle x^2 + x + 1 \;=\;A(x^2+2x+2) + Bx(x+1) + C(x+1)\)
\(\displaystyle \text{Let }x = -1\!: \;\;1 \:=\:A(1) + B(0) + C(0) \quad\Rightarrow\quad A \:=\:1\)
\(\displaystyle \text{Let }x = 1\!:\;\;3 \:=\:5A + 2B + 2C \quad\Rightarrow\quad B + C \:=\:-1\) .[1]
\(\displaystyle \text{Let }x = 2\!:\;\;7 \:=\:10 + 6B + 3C \quad\Rightarrow\quad 2B + C \:=\:-1\) .[2]
Subtract [2] - [1]: .\(\displaystyle B = 0 \quad\Rightarrow \quad C = -1\)
\(\displaystyle \text{The integral becomes: }\;\int^1_0\left(\frac{1}{x+1} - \frac{1}{x^2+2x+2}\right)dx \;\;=\;\;\int^1_0\!\frac{dx}{x+1} \;-\; \int^1_0\frac{dx}{(x+1)^2+1}\)
. . . . . . . . . . . . . . \(\displaystyle =\;\; \ln(x+1)\;-\;\arctan(x+1)\: \bigg]^1_0 \;\;=\;\;\bigg[\ln(2) - \arctan(2)\bigg] - \bigg[\ln(1) - \arctan(1)\bigg]\)
. . . . . . . . . . . . . . \(\displaystyle =\;\;\ln(2) - \arctan(2) + \frac{\pi}{4} \;\;\approx\;\; 0.3714\)
The second one uses Partial Fractions . . .
\(\displaystyle \int_{0}^{1}\frac{x^2+x+1}{(x^2+2x+2)(x+1)}\,dx\)
\(\displaystyle \text{We have: }\;\frac{x^2+x+1}{(x+1)(x^2+2x+2)} \;=\;\frac{A}{x+1} + \frac{Bx}{x^2+2c+2} + \frac{C}{x^2+2x+2}\)
. . . . . . . . . . . . . . \(\displaystyle x^2 + x + 1 \;=\;A(x^2+2x+2) + Bx(x+1) + C(x+1)\)
\(\displaystyle \text{Let }x = -1\!: \;\;1 \:=\:A(1) + B(0) + C(0) \quad\Rightarrow\quad A \:=\:1\)
\(\displaystyle \text{Let }x = 1\!:\;\;3 \:=\:5A + 2B + 2C \quad\Rightarrow\quad B + C \:=\:-1\) .[1]
\(\displaystyle \text{Let }x = 2\!:\;\;7 \:=\:10 + 6B + 3C \quad\Rightarrow\quad 2B + C \:=\:-1\) .[2]
Subtract [2] - [1]: .\(\displaystyle B = 0 \quad\Rightarrow \quad C = -1\)
\(\displaystyle \text{The integral becomes: }\;\int^1_0\left(\frac{1}{x+1} - \frac{1}{x^2+2x+2}\right)dx \;\;=\;\;\int^1_0\!\frac{dx}{x+1} \;-\; \int^1_0\frac{dx}{(x+1)^2+1}\)
. . . . . . . . . . . . . . \(\displaystyle =\;\; \ln(x+1)\;-\;\arctan(x+1)\: \bigg]^1_0 \;\;=\;\;\bigg[\ln(2) - \arctan(2)\bigg] - \bigg[\ln(1) - \arctan(1)\bigg]\)
. . . . . . . . . . . . . . \(\displaystyle =\;\;\ln(2) - \arctan(2) + \frac{\pi}{4} \;\;\approx\;\; 0.3714\)
Hello, steve.b!
The first one requires a Product-to-Sum identity:
.\(\displaystyle \sin A\cos B \;=\;\frac{1}{2}\bigg[\sin(A-B) + \sin(A+B)\bigg]\)
\(\displaystyle \text{Using the identity, the integral becomes: }\;\int^{\frac{\pi}{2}}_0 \frac{1}{2}\bigg[\sin(2x) + \sin(8x)\bigg]\,dx\;\;=\;\;\frac{1}{2}\int^{\frac{\pi}{2}}_0\bigg[\sin(2x) + \sin(8x)\bigg]\,dx\)
. . \(\displaystyle =\;\; \frac{1}{2}\bigg[-\frac{1}{2}\cos(2x) - \frac{1}{8}\cos(8x)\bigg]^{\frac{\pi}{2}}_0 \;\;=\;\;-\frac{1}{16}\bigg[4\cos(2x) + \cos(8x)\bigg]^{\frac{\pi}{2}}_0\)
. . \(\displaystyle =\;\;-\frac{1}{16}\bigg[4\cos(\pi) \;+\; \cos(4\pi)\bigg] \;+\; \frac{1}{16}\bigg[4\cos(0) \;+\; \cos(0)\bigg] \;\;=\;\;-\frac{1}{16}\bigg[4(-1)\;+\;1\bigg] + \frac{1}{16}\bigg[4(1)\;+\;1\bigg]\)
. . \(\displaystyle =\;\;\frac{3}{16} + \frac{5}{16} \;\;=\;\;\frac{1}{2}\)
The first one requires a Product-to-Sum identity:
.\(\displaystyle \sin A\cos B \;=\;\frac{1}{2}\bigg[\sin(A-B) + \sin(A+B)\bigg]\)
\(\displaystyle \int_{0}^{\pi/2}\sin(5x)\cos(3x)\mathrmb{d}x\)
\(\displaystyle \text{Using the identity, the integral becomes: }\;\int^{\frac{\pi}{2}}_0 \frac{1}{2}\bigg[\sin(2x) + \sin(8x)\bigg]\,dx\;\;=\;\;\frac{1}{2}\int^{\frac{\pi}{2}}_0\bigg[\sin(2x) + \sin(8x)\bigg]\,dx\)
. . \(\displaystyle =\;\; \frac{1}{2}\bigg[-\frac{1}{2}\cos(2x) - \frac{1}{8}\cos(8x)\bigg]^{\frac{\pi}{2}}_0 \;\;=\;\;-\frac{1}{16}\bigg[4\cos(2x) + \cos(8x)\bigg]^{\frac{\pi}{2}}_0\)
. . \(\displaystyle =\;\;-\frac{1}{16}\bigg[4\cos(\pi) \;+\; \cos(4\pi)\bigg] \;+\; \frac{1}{16}\bigg[4\cos(0) \;+\; \cos(0)\bigg] \;\;=\;\;-\frac{1}{16}\bigg[4(-1)\;+\;1\bigg] + \frac{1}{16}\bigg[4(1)\;+\;1\bigg]\)
. . \(\displaystyle =\;\;\frac{3}{16} + \frac{5}{16} \;\;=\;\;\frac{1}{2}\)
soroban said:Hello, steve.b!
The first one requires a Product-to-Sum identity:
.\(\displaystyle \sin A\cos B \;=\;\frac{1}{2}\bigg[\sin(A-B) + \sin(A+B)\bigg]\)
\(\displaystyle \int_{0}^{\pi/2}\sin(5x)\cos(3x)\mathrmb{d}x\)
\(\displaystyle \text{Using the identity, the integral becomes: }\;\int^{\frac{\pi}{2}}_0 \frac{1}{2}\bigg[\sin(2x) + \sin(8x)\bigg]\,dx\;\;=\;\;\frac{1}{2}\int^{\frac{\pi}{2}}_0\bigg[\sin(2x) + \sin(8x)\bigg]\,dx\)
. . \(\displaystyle =\;\; \frac{1}{2}\bigg[-\frac{1}{2}\cos(2x) - \frac{1}{8}\cos(8x)\bigg]^{\frac{\pi}{2}}_0 \;\;=\;\;-\frac{1}{16}\bigg[4\cos(2x) + \cos(8x)\bigg]^{\frac{\pi}{2}}_0\)
. . \(\displaystyle =\;\;-\frac{1}{16}\bigg[4\cos(\pi) \;+\; \cos(4\pi)\bigg] \;+\; \frac{1}{16}\bigg[4\cos(0) \;+\; \cos(0)\bigg] \;\;=\;\;-\frac{1}{16}\bigg[4(-1)\;+\;1\bigg] + \frac{1}{16}\bigg[4(1)\;+\;1\bigg]\)
. . \(\displaystyle =\;\;\frac{3}{16} + \frac{5}{16} \;\;=\;\;\frac{1}{2}\)
Dear Soroban!
I am very grateful for your help.
I really appreciate that you sacrificed your time to me.
Thank you so much!