Integral question $\int x \sqrt {1-x}\ dx.$

Question

So there are (from my knowledge) $2$ ways of solving for $\int x \sqrt {1-x}\ dx.$

The first is by $u$ substitution and the second is by parts. They both differ now i'm confused which one is correct? I have both the correct answers but they differ?

Answer 1

Let me guess: you substituted $u=1-x$ to get $\int(u^{3/2}-u^{1/2})du=\frac{2}{5}(1-x)^{5/2}-\frac{2}{3}(1-x)^{3/2}+C_1$, but for parts you wrote $f=x,\,g=-\frac{2}{3}(1-x)^{3/2}$ to get $$-\frac{2}{3}x(1-x)^{3/2}+\frac{2}{3}\int(1-x)^{3/2}dx=-\frac{2}{3}x(1-x)^{3/2}-\frac{4}{15}(1-x)^{5/2}+C_2.$$But from $x(1-x)^{3/2}=(1-x)^{3/2}-(1-x)^{5/2}$, these results are identical (because $\frac{2}{3}-\frac{4}{15}=\frac{2}{5}$) with $C_1=C_2$. (Sometimes when you calculate an indefinite integral by multiple methods, the integration constants differ, but that's fine.)

Answer 2

Try taking the expression for one of your answers and subtracting from the expression for your other answer.

You should find that you can reduce it to a single constant (which may be zero or non-zero). This is expected. The indefinite integral is always expressed with a "$+c$" at the end of the answer - an arbitrary constant of integration. There is no one answer to an indefinite integral, the answer is an infinite family of expressions, all separated from each other by every conceivable constant.

So both your answers are equivalent and equally correct. Provided you did the steps correctly, of course.

Answer 3

$$x\sqrt{1-x}=(x-1+1)\sqrt{1-x}=-(1-x)\sqrt{1-x}-\sqrt{1-x}$$ $$=-(1-x)^{\frac{3}{2}}-(1-x)^{\frac{1}{2}}$$ The anti derivative of $(1-x)^a$ is $-\frac{1}{a+1}(1-x)^{a+1}, a\ne -1$