U substitution
- Thread starter Lost
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pka
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- Jan 29, 2005
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After doing several problems, this one lost me. Can someone explain this step by step? Thanks.
\(\displaystyle D_x\left[\sin(x^3+9)\right]=3x^2\cos(x^3+9)\)
Hello, Lost!
Paraphrasing pka's approach . . .
We have: .\(\displaystyle \displaystyle\int \left[\sin(x^3+9)\right]^{\frac{1}{2}}\left[x^2\cos(x^3+9)\,dx\right] \)
Let \(\displaystyle u \,=\,\sin(x^3+9)\)
n.\(\displaystyle du \,=\,\cos(x^3+9)\cdot 3x^2\,dx \quad\Rightarrow\quad x^2\cos(x^3+9)\,dx \:=\:\frac{1}{3}du\)
Substitute: .\(\displaystyle \displaystyle \int u^{\frac{1}{2}}\left(\tfrac{1}{3}du\right) \;=\;\tfrac{1}{3}\int u^{\frac{1}{2}}du\)
Got it?
Paraphrasing pka's approach . . .
We have: .\(\displaystyle \displaystyle\int \left[\sin(x^3+9)\right]^{\frac{1}{2}}\left[x^2\cos(x^3+9)\,dx\right] \)
Let \(\displaystyle u \,=\,\sin(x^3+9)\)
n.\(\displaystyle du \,=\,\cos(x^3+9)\cdot 3x^2\,dx \quad\Rightarrow\quad x^2\cos(x^3+9)\,dx \:=\:\frac{1}{3}du\)
Substitute: .\(\displaystyle \displaystyle \int u^{\frac{1}{2}}\left(\tfrac{1}{3}du\right) \;=\;\tfrac{1}{3}\int u^{\frac{1}{2}}du\)
Got it?
Thanks Soroban for paraphrasing. My mind works very visually, and pka's response just really didn't help me understand. I know that different people work and look at things differently. For me, a step by step guide is much more helpful.
I do have one question though.
How did you get x^3?
I do have one question though.
How did you get x^3?
Because you told us so. Please look at your original post.