using trig sub vs u-sub with $ \int_0^1 x^3 \sqrt{1 - x^2} \, dx $?

Question

I'm having a hard time understanding when I should use $u$-sub rather than trig sub. I have the problem $$ \int_0^1 x^3 \sqrt{1 - x^2} \, dx $$

My initial attempt was to use $x = \sin u$ as my trig sub but this was wrong. I didn't think to use $u$-sub as the $dx$ and $du$ wouldn't cancel out nicely. But the answer says to just use the $u$ equation and solve for $x$ then plug that into the $x^3$.

This brings me to my next question. If I have an square root integral and both $a$ and $x$ are $1$, do I need to use $u$-sub rather than trig sub?

Answer 1

This is a good exercise because the answer is do both!

The "Pythagorean" expression (a square-root of a sum or difference of squares) suggests a trigonometric substitution. With $x = \sin t$, we have $\sqrt{1 - x^2} = \cos t$ (on the domain $0 \leq x \leq 1$ which corresponds to $0 \leq t \leq \pi/2$). Also, by differentiating, $dx = \cos t \, dt$. Putting this all together, $$ \int_0^1 x^3 \sqrt{1 - x^2} \, dx = \int_0^{\pi/2} \sin^3 t \, \cos^2 t \, dt $$

Now, the standard way to approach this integral is to recognize the odd power of $\sin t$, which suggests peeling off one copy to stick to the differential $dt$, converting the remaining integrand into a polynomial in $\cos t$. Try it yourself before revealing:

$$ \int_0^{\pi/2} \sin^3 t \, \cos^2 t \, dt = \int_0^{\pi/2} \sin^2 t \, \cos^2 t \, (\sin t \, dt) = \int_0^{\pi/2} (1 - \cos^2 t) \, \cos^2 t \, (\sin t \, dt). $$

Now, we make the substitution $u = \cos t$, which has differential $du = -\sin t \, dt$, so we are calculating

$$ - \int_1^0 (1 - u^2) \, u^2 \, du = \int_0^1 (u^2 - u^4) \, du $$

Finally, we evaluate

$$ \int_0^1 (u^2 - u^4) \, du = \bigl(\tfrac13 u^3 - \tfrac15 u^5 \bigr)\bigg|_0^1 = \tfrac13 - \tfrac15 = \tfrac{2}{15}. $$

---

If you just want the value of the definite integral, then you're done. But, this example can teach you something more subtle. Whenever you make two substitutions, you can go back and redo the calculation with a single substitution that does both at once. And when you do this, your integrand goes from algebraic to trigonometric and back to algebraic. $$ u = \cos t = \cos (\arcsin x) = \sqrt{1 - x^2} $$ This suggests that the substitution $u = \sqrt{1 - x^2}$ would have worked from the start. Let's try: $u^2 = 1 - x^2$, so $2u \, du = - 2x \, dx$ for the differentials. Also, $x^2 = 1 - u^2$, and for the the limits, the $0$ and $1$ swap. Thus,

$$ \int_0^1 x^3 \sqrt{1 - x^2} \, dx = \int_0^1 x^2 \sqrt{1 - x^2} (x \, dx) = \int_1^0 (1 - u^2) \, u \, (-u \, du) = \int_0^1 (u^2 - u^4) \, du $$

as before, but with no explicit reference to any trigonometric function.

Answer 2

Letting $u=x^2$ transforms the integral into $$ \begin{aligned} \int_0^1 x^3 \sqrt{1-x^2} d x &=\int_0^1 x^2 \sqrt{1-x^2} \cdot x d x \\ &=\frac{1}{2} \int_0^1 u \sqrt{1-u} d u \end{aligned} $$ Decomposing the integrand into two yields $$ \begin{aligned} I &=\frac{1}{2} \int_0^1[1-(1-u)] \sqrt{1-u} d u \\ &=\frac{1}{2} \int_0^1 \sqrt{1-u} d u-\int_0^1(1-u)^{\frac{3}{2}} d u \\ &=\frac{1}{2}\left[-\frac{2}{3}(1-u)^{\frac{3}{2}}+\frac{2}{5}(1-u)^{\frac{2}{2}}\right]_0^1\\&=\frac{2}{15} \end{aligned} $$

---

For fun, we can do it by integration by parts. $$ \begin{aligned} I &=-\frac{1}{3} \int_0^1 x^2 d\left(1-x^2\right)^{\frac{1}{2}} \\ &=-\left[\frac{x^2\left(1-x^2\right)^{\frac{3}{2}}}{3}\right]_0^1+\frac{2}{3} \int_0^1 x\left(1-x^2\right)^{\frac{3}{2}} d x \\ &=-\frac{2}{3}\left[\frac{1}{5}\left(1-x^2\right)^{\frac{5}{2}}\right]_0^1=\frac{2}{15} \end{aligned} $$