# Thread: [SOLVED] Integration by substitution

1. ## [SOLVED] Integration by substitution

Use the substitution to evaluate

The problem before this was using the same substitution, but to evaluate instead. I was able to do that one easily, but I don't see how I can use this substitution for this problem.

2. Originally Posted by ban26ana
Use the substitution to evaluate

The problem before this was using the same substitution, but to evaluate instead. I was able to do that one easily, but I don't see how I can use this substitution for this problem.

Observe that $u={\color{blue}1-x^2}\implies {\color{red}x^2}=1-u$, then $\,du={\color{green}-2x\,dx}$. The limits of integration change as well: $u(0)=1$ and $u(1)=0$

So $\int_0^1 x^3\sqrt{1-x^2}\,dx=-\tfrac{1}{2}\int_0^1{\color{green}-2x}\cdot {\color{red}x^2}\sqrt{{\color{blue}1-x^2}}{\color{green}\,dx}$ $\xrightarrow{u=1-x^2}{}-\tfrac{1}{2}\int_1^0\left(1-u\right)\sqrt{u}\,du=\tfrac{1}{2}\int_0^1u^{1/2}-u^{3/2}\,du$

Can you continue from here?

3. Yes I can. Thanks so much!

4. Originally Posted by ban26ana
Use the substitution to evaluate

The problem before this was using the same substitution, but to evaluate instead. I was able to do that one easily, but I don't see how I can use this substitution for this problem.
Good question.
You have the substitution $u=1-x^2$
so $\sqrt{1-x^2}=\sqrt{u}$
The problem here is: you want to find $x^3$ and $dx$ in terms of $u$.
Its easy; solve the substitution for $x$ then differentiate both side with respect to $x$ to get the $dx$.
Also find $x^3$ from your substitution.
Can you do that?