<s>[INST] Definite Integrals problem The question is to find the value of : $$\frac{\displaystyle29\int_0^1 (1-x^4)^7\,dx}{\displaystyle4\int_0^1 (1-x^4)^6\,dx}$$ without expanding. According to the book, the answer is 7.
I tried taking $I=\displaystyle\int_0^1(1-x⁴)^7\,dx$ and integrating it by parts taking 1 as the function to be integrated, but I could not manage the $-4x^3$ terms that I was getting by the chain rule.
I cannot figure out how to get the answer.
 [/INST] For the integration by parts, you apply a trick. You don't take the constant $1$ as the factor, but you calculate
\begin{align}
\int_0^1 (1-x^4)^7\,dx &= \int_0^1 (1-x^4)\cdot(1-x^4)^6\,dx\\
&= \int_0^1 (1-x^4)^6\,dx - \int_0^1 x^4\cdot (1-x^4)^6\,dx\\
&= \int_0^1 (1-x^4)^6\,dx + \frac{1}{28} \int_0^1 x\bigl(7(1-x^4)^6\cdot (-4x^3)\bigr)\,dx\\
&= \int_0^1 (1-x^4)^6\,dx + \frac{1}{28}\biggl(\Bigl[x(1-x^4)^7\Bigr]_0^1 - \int_0^1 (1-x^4)^7\,dx\biggr)\\
&= \int_0^1 (1-x^4)^6\,dx - \frac{1}{28}\int_0^1 (1-x^4)^7\,dx.
\end{align}
Moving the last term to the left and dividing by the factor, we obtain
$$\int_0^1 (1-x^4)^7\,dx = \frac{28}{29} \int_0^1 (1-x^4)^6\,dx,$$
as desired.
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