# Integral involving cube root and seventh root

Find the value of $$\int_{0}^{1} (1-x^7)^{\frac{1}{3}}-(1-x^3)^{\frac{1}{7}}\:dx$$

My Approach:

Let $$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$ and $$I_2=\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dx$$

For $I_1$ substitute $x^7=1-t^3$ so $dx=\frac{-3t^2}{7}(1-t^3)^{\frac{-6}{7}}\:dt$ Hence

$$I_1=\int_{1}^{0}\frac{-3t^3}{7}(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$

$$I_1=\frac{-3}{7}\int_{0}^{1}(1-t^3-1)(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$

$$I_1=\frac{-3I_2}{7}+\frac{3A}{7}$$

where $A=\int_{0}^{1}(1-t^3)^{\frac{-6}{7}}dt$

Similarly using substitution $x^3=1-t^7$ for $I_2$ and proceeding as above we get

$$I_2=\frac{-7I_1}{3}+\frac{7B}{3}$$ where $B=\int_{0}^{1}(1-t^7)^{\frac{-2}{3}}dt$

But we need to find $I_1-I_2$ so got stuck up here

• Thanks all pretty happy to see wonderful methods – Ekaveera Kumar Sharma Sep 21 '15 at 6:31
• A related question. – Lucian Sep 21 '15 at 9:56

## 4 Answers

We can write it as $$I = \displaystyle \int_{0}^{1}\sqrt[3]{1-x^7}dx-\int_{0}^{1}\sqrt[7]{1-x^3}dx$$

Now Using $$\displaystyle \bullet \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$

So we get $$I = \displaystyle \int_{0}^{1}\left(1-x^7\right)^{\frac{1}{3}}dx+\int_{1}^{0}\left(1-x^3\right)^{\frac{1}{7}}dx$$

Now Let $$\displaystyle f(x) = \left(1-x^{7}\right)^{\frac{1}{3}}\;,$$ Then $$f^{-1}(x) = (1-x^3)^{\frac{1}{7}}$$ and also $f(0) = 1$ and $f(1) =0$

So Integral $$\displaystyle I = \int_{0}^{1}f(x)dx+\int_{f(0)}^{f(1)}f^{-1}(x)dx$$

Now let $f^{-1}(x) = z\;,$ Then $x=f(z)$ So we get $dx = f'(z)dz$

So Integral $$\displaystyle I =\int_{0}^{1}f(x)dx+\int_{0}^{1}z\cdot f'(z)dz$$

Now Integration by parts for second Integral, We get

$$\displaystyle I =\int_{0}^{1}f(x)dx+\left[z\cdot f(z)\right]_{0}^{1}-\int_{0}^{1}f(z)dz$$

So using $$\displaystyle \bullet\; \int_{a}^{b}f(z)dz = \int_{a}^{b}f(x)dx$$

So we get $$\displaystyle I =\int_{0}^{1}f(x)dx+f(1) -\int_{0}^{1}f(x)dx = f(1) =0$$

• This is more or less what I posted, but approaches the problem of showing equality a little differently, using inverse functions... I like it! – Brevan Ellefsen Sep 21 '15 at 6:23
• Have you missed a minus in your second equation (the one with bullet)? – Ruslan Sep 21 '15 at 13:51
• Thanks Ruslan i have edited it. – juantheron Sep 21 '15 at 16:06

Hint. A straightforward approach is to use the Euler beta function $$B(a,b) := \int _0^1 x^{a-1}(1-x)^{b-1}dx =\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},\quad a>0, b>0.$$ Then you get $$\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dt-\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dt=0.$$

Consider the area in the first quadrant bounded by the curve $x^7+y^3 = 1$.

The curve in the first quadrant can be written as $y = (1-x^7)^{1/3}$ for $0 \le x \le 1$. Hence, the area bounded by the curve is given by $\displaystyle\int_{0}^{1}(1-x^7)^{1/3}\,dx$.

We can also write the curve in the first quadrant as $x = (1-y^3)^{1/7}$ for $0 \le y \le 1$. So, the area bounded by the curve is also $\displaystyle\int_{0}^{1}(1-y^3)^{1/7}\,dy$.

Therefore, $\displaystyle\int_{0}^{1}(1-x^7)^{1/3}\,dx = \int_{0}^{1}(1-y^3)^{1/7}\,dy$, and thus, the difference is $\displaystyle\int_{0}^{1}\left[(1-x^7)^{1/3} - (1-x^3)^{1/7}\right]\,dx = 0$.

$$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$ substitute $t = 1-x^7$, $\,\,dt = -7x^6dx$, and by extension $\,\,x = \sqrt[7]{1-t}$ $$=-\frac{1}{7}\int_{0}^{1} \frac{\sqrt[3]{t}}{(1-t)^\frac{6}{7}}dx$$ Now integrate by parts choosing the factors $u = \sqrt[3]{t}$ $\,\, u' = \frac{1}{3t^{2/3}}$$\,\,dv = \frac{1}{(1-t)^{6/7}}\,\, and v = -7(1-t)^{1/7}$$ = -\frac{1}{7}(-\sqrt[3]{t}\sqrt[7]{1-t})\bigg|_0^1 -\frac{1}{7}\int_0^1 \frac{-7\sqrt[7]{1-t}}{3t^{2/3}}dt = -\frac{1}{7}\int_0^1 \frac{-7\sqrt[7]{1-t}}{3t^{2/3}}dt$$Finally, subsitute t = v^3, \,\, dt = 3v^2dv$$ = \int_0^1 \sqrt[7]{1-v^3}dv = I_2$$So what we have is$$\int_{0}^{1} (I_2 - I_2)=\int_{0}^{1} 0=0$\$