+267 Hello, I tried posting this question yesterday but it must have gotten caught in the spam filter because I included a link, I'll avoid doing it this time, since I'm too lazy to rewrite it I'll be brief. As the title said I tried I'm having trouble integrating:
∫(2xsin3x)dx
I first moved the 2 outside the integral and tried applying integration by parts which lead me to the following:
6xcos3(x)−18xcos(x)−2sin3(x)−12sin(x)9
I chose x to be equal to u as to derivate it and get rid of it when I'd have to integrate the second time. I ended up having to integrate sin^3(x) and cos^3(x) if I remember correctly. I did this question twice with the same result.
After looking the answer up on some online integral calculating site it gave something else. I compared the two to see if they were perhaps equal but written in another form but without success.
I'd appreciate a step by step solution or a link to one. That'd probably clear things up as I can't be able to see the error I made, thanks alot!
integral2 x sin^3(x) dx = 1/18 (27 sin(x)-sin(3 x)-27 x cos(x)+3 x cos(3 x))+constant
+26396 integrating:
∫dx 2xsin3(x)
1. Let
∫dx sin3(x)=∫dx sin2(x)⋅sin(x)=∫dx [1−cos2(x)]⋅sin(x)u=cos(x)du=−sin(x)dx=∫du 1−sin(x)⋅(1−u2)⋅sin(x)=−∫du (1−u2)=−∫du+∫du u2=−u+13⋅u3=−cos(x)+13⋅cos3(x)
2. Let
∫dx cos3(x)=∫dx cos2(x)⋅cos(x)=∫dx [1−sin2(x)]⋅cos(x)u=sin(x)du=cos(x)dx=∫du 1cos(x)⋅(1−u2)⋅cos(x)=∫du (1−u2)=∫du−∫du u2=u−13⋅u3=sin(x)−13⋅sin3(x)
3. Let
∫dx x⋅sin3(x)u=xu′=1v=−cos(x)+13⋅cos3(x)v′=sin3(x)=x⋅[−cos(x)+13⋅cos3(x)]−∫dx 1⋅[−cos(x)+13⋅cos3(x)]=−x⋅cos(x)+13⋅x⋅cos3(x)+∫dx cos(x)−∫dx 13⋅cos3(x)=−x⋅cos(x)+13⋅x⋅cos3(x)+sin(x)−13⋅∫dx cos3(x)=−x⋅cos(x)+13⋅x⋅cos3(x)+sin(x)−13⋅[sin(x)−13⋅sin3(x)]=−x⋅cos(x)+13⋅x⋅cos3(x)+sin(x)−13⋅sin(x)+19⋅sin3(x)]=−x⋅cos(x)+13⋅x⋅cos3(x)+23⋅sin(x)+19⋅sin3(x)∫dx 2⋅x⋅sin3(x)=−2x⋅cos(x)+23⋅x⋅cos3(x)+43⋅sin(x)+29⋅sin3(x)=19⋅[−18x⋅cos(x)+6x⋅cos3(x)+12⋅sin(x)+2⋅sin3(x)]
4. Let
cos3(x)=14⋅[3⋅cos(x)+cos(3x)]sin3(x)=14⋅[3⋅sin(x)−sin(3x)]
5. Let
∫dx 2⋅x⋅sin3(x)=19⋅[−18x⋅cos(x)+6x⋅cos3(x)+12⋅sin(x)+2⋅sin3(x)]=19⋅{−18x⋅cos(x)+6x⋅14⋅[3⋅cos(x)+cos(3x)]+12⋅sin(x)+2⋅14⋅[3⋅sin(x)−sin(3x)]}=−2x⋅cos(x)+16⋅x⋅[3⋅cos(x)+cos(3x)]+43⋅sin(x)+118⋅[3⋅sin(x)−sin(3x)]=−2x⋅cos(x)+12⋅x⋅cos(x)+16⋅x⋅cos(3x)+43⋅sin(x)+16⋅sin(x)−118⋅sin(3x)=−32⋅x⋅cos(x)+16⋅x⋅cos(3x)+32⋅sin(x)−118⋅sin(3x)
or
∫dx 2⋅x⋅sin3(x)=−32⋅x⋅cos(x)+16⋅x⋅cos(3x)+32⋅sin(x)−118⋅sin(3x)=118[−27⋅x⋅cos(x)+3⋅x⋅cos(3x)+27⋅sin(x)−sin(3x)]
