<p>Hey guys my teacher is evil and gave us these problems to do for the weekend. I am having trouble. Please help</p>
<p>Use integration by parts to evaluate integral x sin (2x - 5) dx.</p>
<p>Solve the differential equation by separation of variables.
xcos(x^2)</p>
<hr>
<p>(y^2)-2</p>
<p>and </p>
<p>Use substitution to evaluate the integral.</p>
<p>rad (tan x^5)secx^2 dx</p>
<p>Can you guys please show step by step how you did it? I would be greatful</p>
<p>Bumppppppp</p>
<p>i'll help with that first integration by parts one:</p>
<p>let u= x so du=dx
let dv= sin(2x-5) so v= -1/2cos(2x-5)</p>
<p>use the formula uv - integral of(v*du)</p>
<p>you get
(-1/2x)cos(2x-5) - int(-1/2cos(2x-5) dx)</p>
<p>final answer
(-1/2x)(cos(2x-5)) - (1/4)(sin(2x-5))</p>
<p>and i think that's the answer. if i'm wrong, oops.</p>
<p>thank you so much dvlfnfv5 I really appreciate it. Can others please help me with the other two? I really appreciate it once agian. Thank you.</p>
<p>Hey here's the separable equation step-by-step:</p>
<p>dy/dx=(xcos(x^2))/(y^2-2) so cross multiply</p>
<p>(y^2-2)dy=xcos(x^2)dx then integrate</p>
<p>the xcosx^2 is the tricky part, you have to substitute u=x^2, so du=2xdx, and dx=du/2x. Then plug that in for dx, and you get intcosu, which is simple. Then integrate, and it becomes </p>
<p>y^3-2y=sinu</p>
<p>y^3-2y=sin(x^2) Then just rearrange for y</p>
<p>Hope that helps, and that I didn't make mistakes...</p>
<p>this def helps. Thank you so much bre! Can someone please help me with the last problem? THank you so much. I relaly appreciate it.</p>
<p>There's a definite trick to that last one, like making u=something or another...sorry I'm not much more help. I'm sure that was a homework problem, but we finished doing calc a couple weeks ago...
I'll go check see if I can find anything to help with the last one.</p>
<p>omg that would be amazing bre. Thank you so much.</p>
<p>last problem:
let u = tanx, du = secx^2
so its integral of u^5 = u6 over 6 so its tanx^6 over 6</p>
<p>Wow nicely done.</p>
<p>Wait, but its tan(x^5), not (tanx)^5...</p>
<p>What's the "rad" in front of the problem? </p>
<p>like radical? or something</p>
<p>well on the sheet it says rad (tan^5 x) so isnt that tanx ^5?</p>
<p>Sorry, I couldn't find anything, but I think its more complex than neoa's solution...I think there's a different substitution.</p>
<p>yea its radical</p>
<p>Oh, I don't know what rad means.</p>
<p>yea I thought so too because of the rad, however if there wasnt the rad, it would be the right answer I presume</p>
<p>lol sry bre, I wasnt telling you, I was telling the other poster who asked what rad was. My apologies</p>