Again, Sorry, but need help on Evaluating the integral
- Thread starter femmed0ll
- Start date
D
Deleted member 4993
Guest
femmed0ll said:
I have no idea - how I could help you here. Only advice I have for you to go back and "take trig". Otherwise this is going to remain very confusing.
Whoever advised you to take calculus without trigonometry - should be hung by "its" toes....
femmed0ll said:
They give you the sub to make. Since \(\displaystyle u=x^{\frac{3}{2}}-7, \;\ du=\frac{3}{2}\sqrt{x}dx, \;\ \frac{2}{3}du=\sqrt{x}dx\)
See?. Upon making the subs above, you get \(\displaystyle \frac{2}{3}\int cos^{2}(u)du\)
Now, integrate and resub.
The trick to integrating \(\displaystyle cos^{2}(u)\) is to use the identity \(\displaystyle cos^{2}(u)=\frac{cos(2u)+1}{2}\)
Since \(\displaystyle u=cot(7\theta), \;\ du=-7csc^{2}(7\theta)d{\theta}, \;\ \frac{-du}{7}=csc^{2}(7\theta)d{\theta}\)
See it there?. Makes the subs, integrate, then resub.
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
femmed0ll said:
What ? ? ! :?
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle First \ one, \ in \ its \ entirety.\)
\(\displaystyle \int \sqrt x cos^2(x^{3/2}-7)dx.\)
\(\displaystyle Let \ u \ = \ x^{3/2}-7, \ then \ \frac{2}{3}du \ = \ x^{1/2}dx.\)
\(\displaystyle Ergo, \ we \ now \ have: \ \frac{2}{3}\int cos^2(u)du, \ not \ too \ bad.\)
\(\displaystyle Therefore, \ applying \ an \ identity \ gives \ us \ \frac{1}{3}\int\ [1+cos(2u)]du.\)
\(\displaystyle Now, \ this \ reduces \ to: \ \frac{1}{3}\bigg[u+\frac{sin(2u)}{2}\bigg]+C_1.\)
\(\displaystyle Using \ another \ identity \ gives \ us: \ \frac{1}{3}\bigg[u+sin(u)cos(u)\bigg]+C_1.\)
\(\displaystyle Now, \ resubbing \ gives \ us: \ \frac{1}{3}\bigg[x^{3/2}-7+sin(x^{3/2}-7)cos(x^{3/2}-7)\bigg]+C_1.\)
\(\displaystyle Finally, \ we \ have: \ \frac{1}{3}\bigg[x^{3/2}+sin(x^{3/2}-7)cos(x^{3/2}-7)\bigg]+C, \ C \ = \ -\frac{7}{3}+C_1.\)
\(\displaystyle But \ how \ do \ we \ know \ if \ we \ haven't \ messed \ up \ somewhere \ along \ the \ line?\)
\(\displaystyle Ergo, \ preform \ a \ check:\)
\(\displaystyle D_x \bigg[\frac{1}{3}[x^{3/2}+sin(x^{3/2}-7)cos(x^{3/2}-7)]+C\bigg]\)
\(\displaystyle = \ \frac{1}{3}\bigg[\frac{3}{2}x^{1/2}+\frac{3}{2}x^{1/2}cos(x^{3/2}-7)cos(x^{3/2}-7)+\frac{3}{2}x^{1/2}sin(x^{3/2}-7)(-sin(x^{3/2}-7))\bigg]\)
\(\displaystyle = \ \frac{x^{1/2}}{2}[1+cos^2(x^{3/2}-7)-sin^2(x^{3/2}-7)]\)
\(\displaystyle = \ \frac{x^{1/2}}{2}[1+cos^2(x^{3/2}-7)-(1-cos^2(x^{3/2}-7))]\)
\(\displaystyle = \ \frac{x^{1/2}}{2}[2cos^2(x^{3/2}-7)] \ = \ \sqrt x cos^2(x^{3/2}-7), \ QED\)
\(\displaystyle \int \sqrt x cos^2(x^{3/2}-7)dx.\)
\(\displaystyle Let \ u \ = \ x^{3/2}-7, \ then \ \frac{2}{3}du \ = \ x^{1/2}dx.\)
\(\displaystyle Ergo, \ we \ now \ have: \ \frac{2}{3}\int cos^2(u)du, \ not \ too \ bad.\)
\(\displaystyle Therefore, \ applying \ an \ identity \ gives \ us \ \frac{1}{3}\int\ [1+cos(2u)]du.\)
\(\displaystyle Now, \ this \ reduces \ to: \ \frac{1}{3}\bigg[u+\frac{sin(2u)}{2}\bigg]+C_1.\)
\(\displaystyle Using \ another \ identity \ gives \ us: \ \frac{1}{3}\bigg[u+sin(u)cos(u)\bigg]+C_1.\)
\(\displaystyle Now, \ resubbing \ gives \ us: \ \frac{1}{3}\bigg[x^{3/2}-7+sin(x^{3/2}-7)cos(x^{3/2}-7)\bigg]+C_1.\)
\(\displaystyle Finally, \ we \ have: \ \frac{1}{3}\bigg[x^{3/2}+sin(x^{3/2}-7)cos(x^{3/2}-7)\bigg]+C, \ C \ = \ -\frac{7}{3}+C_1.\)
\(\displaystyle But \ how \ do \ we \ know \ if \ we \ haven't \ messed \ up \ somewhere \ along \ the \ line?\)
\(\displaystyle Ergo, \ preform \ a \ check:\)
\(\displaystyle D_x \bigg[\frac{1}{3}[x^{3/2}+sin(x^{3/2}-7)cos(x^{3/2}-7)]+C\bigg]\)
\(\displaystyle = \ \frac{1}{3}\bigg[\frac{3}{2}x^{1/2}+\frac{3}{2}x^{1/2}cos(x^{3/2}-7)cos(x^{3/2}-7)+\frac{3}{2}x^{1/2}sin(x^{3/2}-7)(-sin(x^{3/2}-7))\bigg]\)
\(\displaystyle = \ \frac{x^{1/2}}{2}[1+cos^2(x^{3/2}-7)-sin^2(x^{3/2}-7)]\)
\(\displaystyle = \ \frac{x^{1/2}}{2}[1+cos^2(x^{3/2}-7)-(1-cos^2(x^{3/2}-7))]\)
\(\displaystyle = \ \frac{x^{1/2}}{2}[2cos^2(x^{3/2}-7)] \ = \ \sqrt x cos^2(x^{3/2}-7), \ QED\)