Trig Substitutions I dont know what im doing wrong

[kluda06]

I need help solving

3x³ / √1-x² dx Answer is -(x²+2)√1-x² +C

This is what i did. First I found what cos and sin were.

√1-x²= cos Θ

x= sin Θ
dx= cos Θ dΘ

Now I plugged them in.

3(sinΘ)³ cosΘ dΘ / cosΘ

I cancelled the cosΘ. Now i have

3(sinΘ)³ dΘ
= 3

(sinΘ)³ dΘ
= 3

(sinΘ)² (sinΘ) dΘ
= 3

(1-cos²Θ)(sinΘ)dΘ
= 3

sinΘ - (cos²Θ sinΘ) dΘ
= -3cosΘ-3

(cos²Θ sinΘ) dΘ

I tried using the U substitutions but I keep getting

-3cosΘ+(cosΘ)³+C

When i plug in what cosΘ is i get

-3√1-x² + (√1-x²)³+C

and that is not the answer :/

Can someone help me out with this please?! its driving me crazy!
Also, is there any other way besides using U substitution?

Thank You

 

[stapel]

I need help solving

\(\displaystyle \displaystyle{\int}\, \dfrac{3x^3}{\sqrt{1\, -\, x^2}}\, dx\)

Try using \(\displaystyle u\, =\, 1\, -\, x^2\).

 

[Deleted member 4993]

I need help solving

3x³ / √1-x² dx Answer is -(x²+2)√1-x² +C

This is what i did. First I found what cos and sin were.

√1-x²= cos Θ

x= sin Θ
dx= cos Θ dΘ

Now I plugged them in.

3(sinΘ)³ cosΘ dΘ / cosΘ

I cancelled the cosΘ. Now i have

3(sinΘ)³ dΘ
= 3

(sinΘ)³ dΘ
= 3

(sinΘ)² (sinΘ) dΘ
= 3

(1-cos²Θ)(sinΘ)dΘ
= 3

sinΘ - (cos²Θ sinΘ) dΘ
= -3cosΘ-3

(cos²Θ sinΘ) dΘ

I tried using the U substitutions but I keep getting

-3cosΘ+(cosΘ)³+C

When i plug in what cosΘ is i get

-3√1-x² + (√1-x²)³+C

Just a bit more work!!!

\(\displaystyle \displaystyle = \ -3\sqrt{1-x^2} \ + \ (1-x^2)\sqrt{1-x^2}\)

\(\displaystyle \displaystyle = \ -2\sqrt{1-x^2} \ - \ x^2\sqrt{1-x^2}\)

\(\displaystyle \displaystyle = \ -\sqrt{1-x^2} (2\ + \ x^2)\)

and that is the answer :/

and that is not the answer :/

Can someone help me out with this please?! its driving me crazy!
Also, is there any other way besides using U substitution?

Thank You

.

 

[kluda06]

Thank you!!!!!!

.

thank you!!!! ugh im just so annoyed that i was so blinded by something easy!!!

Thank you thank you!