Again, Sorry, but need help on Evaluating the integral

[femmed0ll]

39) ? ?x cos^2 (x^(3/2) - 7) dx, u = x^(3/2) -7

40) csc^(2) 7?cot 7? d? , u = cot7?

i never took trig, so this is super-confusing...no idea where to start!
can you have a step by step process on how to figure out the answer? thanks!

 

[Deleted member 4993]

39) ? ?x cos^2 (x^(3/2) - 7) dx, u = x^(3/2) -7

40) csc^(2) 7?cot 7? d? , u = cot7?

i never took trig, so this is super-confusing...no idea where to start!
can you have a step by step process on how to figure out the answer? thanks!

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]

reason why i didnt take trig because i took a placement test (last year) and i scored high enough to skip trig and go on to pre-cal....and now im in calculus.
i have a test tomorrow and i just need help.

 

[galactus]

39) $\displaystyle \int \sqrt{x} cos^{2}(x^{3/2} - 7) dx, \;\ u = x^{3/2} -7$

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 $\displaystyle cos^{2}(u)=\frac{cos(2u)+1}{2}$

40) $\displaystyle \int csc^{2}(7\theta)cot(7\theta)d{\theta}, \;\ u = cot(7\theta)$

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]

i never took trig

i didnt take trig because i took a placement test and i scored high enough to skip trig

i just need help [with trigonometry]

What ? ? ! :?

 

[BigGlenntheHeavy]

$\displaystyle First \ one, \ in \ its \ entirety.$

$\displaystyle \int \sqrt x cos^2(x^{3/2}-7)dx.$

$\displaystyle Let \ u \ = \ x^{3/2}-7, \ then \ \frac{2}{3}du \ = \ x^{1/2}dx.$

$\displaystyle Ergo, \ we \ now \ have: \ \frac{2}{3}\int cos^2(u)du, \ not \ too \ bad.$

$\displaystyle Therefore, \ applying \ an \ identity \ gives \ us \ \frac{1}{3}\int\ [1+cos(2u)]du.$

$\displaystyle Now, \ this \ reduces \ to: \ \frac{1}{3}\bigg[u+\frac{sin(2u)}{2}\bigg]+C_1.$

$\displaystyle Using \ another \ identity \ gives \ us: \ \frac{1}{3}\bigg[u+sin(u)cos(u)\bigg]+C_1.$

$\displaystyle Now, \ resubbing \ gives \ us: \ \frac{1}{3}\bigg[x^{3/2}-7+sin(x^{3/2}-7)cos(x^{3/2}-7)\bigg]+C_1.$

$\displaystyle Finally, \ we \ have: \ \frac{1}{3}\bigg[x^{3/2}+sin(x^{3/2}-7)cos(x^{3/2}-7)\bigg]+C, \ C \ = \ -\frac{7}{3}+C_1.$

$\displaystyle But \ how \ do \ we \ know \ if \ we \ haven't \ messed \ up \ somewhere \ along \ the \ line?$

$\displaystyle Ergo, \ preform \ a \ check:$

$\displaystyle D_x \bigg[\frac{1}{3}[x^{3/2}+sin(x^{3/2}-7)cos(x^{3/2}-7)]+C\bigg]$

$\displaystyle = \ \frac{1}{3}\bigg[\frac{3}{2}x^{1/2}+\frac{3}{2}x^{1/2}cos(x^{3/2}-7)cos(x^{3/2}-7)+\frac{3}{2}x^{1/2}sin(x^{3/2}-7)(-sin(x^{3/2}-7))\bigg]$

$\displaystyle = \ \frac{x^{1/2}}{2}[1+cos^2(x^{3/2}-7)-sin^2(x^{3/2}-7)]$

$\displaystyle = \ \frac{x^{1/2}}{2}[1+cos^2(x^{3/2}-7)-(1-cos^2(x^{3/2}-7))]$

$\displaystyle = \ \frac{x^{1/2}}{2}[2cos^2(x^{3/2}-7)] \ = \ \sqrt x cos^2(x^{3/2}-7), \ QED$