Definite Integration Question 51
Question 51
- The value(s) of $\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x$ is (are)
(2010) (a) $\frac{22}{7}-\pi$ (b) $\frac{2}{105}$ (c) 0 (d) $\frac{71}{15}-\frac{3 \pi}{2}$
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Solution:
- Let $I=\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=\int_{0}^{1} \frac{\left(x^{4}-1\right)(1-x)^{4}+(1-x)^{4}}{\left(1+x^{2}\right)} d x$
$$ \begin{aligned} & =\int_{0}^{1}\left(x^{2}-1\right)(1-x)^{4} d x+\int_{0}^{1} \frac{\left(1+x^{2}-2 x\right)^{2}}{\left(1+x^{2}\right)} d x \ & =\int_{0}^{1}\left(x^{2}-1\right)(1-x)^{4}+\left(1+x^{2}\right)-4 x+\frac{4 x^{2}}{\left(1+x^{2}\right)} d x \ & =\int_{0}^{1}\left(x^{2}-1\right)(1-x)^{4}+\left(1+x^{2}\right)-4 x+4-\frac{4}{1+x^{2}} d x \ & =\int_{0}^{1} x^{6}-4 x^{5}+5 x^{4}-4 x^{2}+4-\frac{4}{1+x^{2}} d x \ & =\frac{x^{7}}{7}-\frac{4 x^{6}}{6}+\frac{5 x^{5}}{5}-\frac{4 x^{3}}{3}+4 x-4 \tan ^{-1} x \ & =\frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4 \frac{\pi}{4}-0=\frac{22}{7}-\pi \end{aligned} $$
