## TransendentialPI Group Title I can't see the substitution I'm supposed to make. Any hints? See next for the exercise. 2 years ago 2 years ago

1. TransendentialPI Group Title

2. cwrw238 Group Title

try u = sqrt(2x)

3. TransendentialPI Group Title

I did |dw:1343307585147:dw||dw:1343307655512:dw| I'm supposed to be geting this into the form so I can find the formula in a integral table.

4. TransendentialPI Group Title

I did find this one, but it is making the integral more complicated:

5. TransendentialPI Group Title

That is not getting me to one of those MC answers/

6. waterineyes Group Title

Now use Integration By parts..

7. waterineyes Group Title

$\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1 - x^2}}$

8. waterineyes Group Title

$\int\limits u \cdot \sin^{-1}(u) du = \sin^{-1}(u) \cdot \frac{u^2}{2} - \int\limits( \frac{1}{\sqrt{1 - u^2}} \times \frac{u^2}{2}) du$ Try to solve for last part..

9. waterineyes Group Title

you can do it like this: $\frac{1}{\sqrt{1 - u^2}} \times u^2 = \frac{1 - u^2 + 1}{\sqrt{1 - u^2}} =\sqrt{1 - u^2} + \frac{1}{\sqrt{1 - u^2}}$ Use the following two Special Integral Formula: $\int\limits \sqrt{a^2 - x^2}dx = \frac{1}{2}x \cdot \sqrt{a^2 - x^2} + \frac{1}{2}a^2 \cdot \sin^{-1} (\frac{x}{a}) + C$ $\int\limits \frac{1}{\sqrt{a^2 - u^2}}dx = \sin^{-1}(\frac{x}{a}) + C$

10. waterineyes Group Title

Sorry - will be there in between terms: $\frac{1}{\sqrt{1 - u^2}} \times u^2 = \frac{1 - u^2 - 1}{\sqrt{1 - u^2}} =\sqrt{1 - u^2} - \frac{1}{\sqrt{1 - u^2}}$ By this your integration middle term will become positive: $\int\limits\limits u \cdot \sin^{-1}(u) du = \sin^{-1}(u) \cdot \frac{u^2}{2} + \int\limits\limits( \frac{1}{\sqrt{1 - u^2}} \times \frac{-u^2}{2}) du$

11. TransendentialPI Group Title

Thanks, I just flipped back in the textbook section and they provide an example in the form $\int\limits xsin^{-1} x$ dx= that integral and then to use a the table again. Thank so much!

12. waterineyes Group Title

Welcome dear..