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Q.
$\int_{0}^{1} \frac{x}{(1-x)^{3 / 4}} d x$ is equal to:
Bihar CECEBihar CECE 2002
Solution:
If the product of two functions is given, we can integrate by using the method of integration by parts.
$
\begin{array}{l}
\text { Let } I=\int_{0}^{1} x(1-x)^{-3 / 4} d x \\
=\frac{x(1-x)^{1 / 4}(-1)}{1 / 4}+\int \frac{(1-x)^{1 / 4}}{1 / 4} d x \\
=\left[4 x(1-x)^{1 / 4}-\frac{4(1-x)^{5 / 4}}{5 / 4}\right]_{0}^{1} \\
=\left[0+0-\left(0-\frac{16}{5}\right)\right] \\
=\frac{16}{5}
\end{array}
$
Alternate Solution:
$
\begin{array}{l}
\text { Let } I=\int_{0}^{1} x(1-x)^{-3 / 4} d x \\
=\int_{0}^{1}(1-x)[1-(1-x)] d x \\
=\int_{0}^{1}(1-x) x^{-3 / 4} d x \\
=\int_{0}^{1}\left(x^{-3 / 4}-x^{1 / 4}\right) d x \\
=\left[\frac{x^{1 / 4}}{1 / 4}-\frac{x^{5 / 4}}{5 / 4}\right]_{0}^{1} \\
=\left[4-\frac{4}{5}\right] \\
=\frac{16}{5}
\end{array}
$