You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
integral math
- Thread starter mathe25
- Start date
D
Deleted member 4993
Guest
View attachment 2620 can u help me with one of these ??? please i need it fot tomorrow
Hint:
cos2x = [1+ cos(2x)]/2
D
Deleted member 4993
Guest
sin ^2 x -???
Hint above should guide you.
Hello, mathe25!
Here's the first one . . .
\(\displaystyle \text{Trig identity: }\:\cos^2\!\theta \:=\:\dfrac{1 + \cos2\theta}{2}\)
\(\displaystyle \displaystyle\text{So we have: }\:I \;=\;\int e^x\left(\frac{1+\cos2x}{2}\right)dx \;=\;\tfrac{1}{2}\int e^x\,dx + \tfrac{1}{2}\int e^x\cos2x\,dx \)
. . . . . . . . . .\(\displaystyle \displaystyle I \;=\;\tfrac{1}{2}e^x + \tfrac{1}{2}\int e^x\cos2x\,dx\;\;\color{red}{[1]}\)
\(\displaystyle \displaystyle\text{Let }\,J \:=\:\int e^x\cos2x\,dx\)
\(\displaystyle \text{By parts: }\:\begin{Bmatrix} u &=& \cos2x && dv &=& e^xdx \\ du &=& \text{-}2\sin2x\,dx && v &=& e^x \end{Bmatrix}\)
\(\displaystyle \displaystyle J \;=\;e^x\cos2x + 2\int e^x\sin2x\,dx\)
\(\displaystyle \text{By parts: }\:\begin{Bmatrix}u &=& \sin2x && dv &=& e^xdx \\ du &=& 2\cos2x\,dx && v &=& e^x \end{Bmatrix}\)
\(\displaystyle \displaystyle J \;=\;e^x\cos2x + 2\left(e^x\sin2x - 2\int e^x\cos2x\,dx\right) \)
\(\displaystyle \displaystyle J \;=\;e^x\cos2x + 2e^x\sin2x -4\underbrace{\int e^x\cos2x\,dx}_{\text{This is }J}\)
\(\displaystyle J \;=\;e^x\cos2x + 2e^x\sin2x - 4J +C \)
\(\displaystyle 5J \;=\;e^x\cos2x + 2e^x\sin2x +C\)
\(\displaystyle J \;=\;\frac{1}{5}e^x\cos2x + \frac{2}{5} e^x\sin2x + C\)
\(\displaystyle \text{Substitute into }\color{red}{[1]}:\;I \;=\;\frac{1}{2}e^x + \frac{1}{2}\left(\frac{1}{5}e^x\cos2x + \frac{2}{5}e^x\sin2x\right) + C\)
. . . . . . . . . . . . . . . \(\displaystyle I \;=\;\frac{1}{2}e^x + \frac{1}{10}e^x\cos2x + \frac{1}{5}e^x\sin2x + C\)
. . . . . . . . . . . . . . . \(\displaystyle I \;=\;\frac{1}{10}e^x(5 + \cos2x + 2\sin2x) + C\)
Here's the first one . . .
\(\displaystyle \displaystyle I \;=\;\int e^x\cos^2\!x\,dx\)
\(\displaystyle \text{Trig identity: }\:\cos^2\!\theta \:=\:\dfrac{1 + \cos2\theta}{2}\)
\(\displaystyle \displaystyle\text{So we have: }\:I \;=\;\int e^x\left(\frac{1+\cos2x}{2}\right)dx \;=\;\tfrac{1}{2}\int e^x\,dx + \tfrac{1}{2}\int e^x\cos2x\,dx \)
. . . . . . . . . .\(\displaystyle \displaystyle I \;=\;\tfrac{1}{2}e^x + \tfrac{1}{2}\int e^x\cos2x\,dx\;\;\color{red}{[1]}\)
\(\displaystyle \displaystyle\text{Let }\,J \:=\:\int e^x\cos2x\,dx\)
\(\displaystyle \text{By parts: }\:\begin{Bmatrix} u &=& \cos2x && dv &=& e^xdx \\ du &=& \text{-}2\sin2x\,dx && v &=& e^x \end{Bmatrix}\)
\(\displaystyle \displaystyle J \;=\;e^x\cos2x + 2\int e^x\sin2x\,dx\)
\(\displaystyle \text{By parts: }\:\begin{Bmatrix}u &=& \sin2x && dv &=& e^xdx \\ du &=& 2\cos2x\,dx && v &=& e^x \end{Bmatrix}\)
\(\displaystyle \displaystyle J \;=\;e^x\cos2x + 2\left(e^x\sin2x - 2\int e^x\cos2x\,dx\right) \)
\(\displaystyle \displaystyle J \;=\;e^x\cos2x + 2e^x\sin2x -4\underbrace{\int e^x\cos2x\,dx}_{\text{This is }J}\)
\(\displaystyle J \;=\;e^x\cos2x + 2e^x\sin2x - 4J +C \)
\(\displaystyle 5J \;=\;e^x\cos2x + 2e^x\sin2x +C\)
\(\displaystyle J \;=\;\frac{1}{5}e^x\cos2x + \frac{2}{5} e^x\sin2x + C\)
\(\displaystyle \text{Substitute into }\color{red}{[1]}:\;I \;=\;\frac{1}{2}e^x + \frac{1}{2}\left(\frac{1}{5}e^x\cos2x + \frac{2}{5}e^x\sin2x\right) + C\)
. . . . . . . . . . . . . . . \(\displaystyle I \;=\;\frac{1}{2}e^x + \frac{1}{10}e^x\cos2x + \frac{1}{5}e^x\sin2x + C\)
. . . . . . . . . . . . . . . \(\displaystyle I \;=\;\frac{1}{10}e^x(5 + \cos2x + 2\sin2x) + C\)
D
Deleted member 4993
Guest
\(\displaystyle I \ = \ \int e^x sin^2(x) dx \ = \ \int e^x [1- cos^2(x)] dx \ \)