Q115 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
$\int_0^1 \frac{x}{(1-x)^{3 / 4}} d x=$
- A$\frac{4}{5}$
- B$\frac{8}{15}$
- C$\frac{14}{5}$
- Correct answerD$\frac{16}{5}$
Solution
$\begin{aligned} & \text { } I=\int_0^1 \frac{x}{(1-x)^{3 / 4}} d x, x=\sin ^2 \theta \Rightarrow d x=2 \sin \theta \cos \theta d \theta \\ & \because I=\int_0^{\pi / 2} \frac{\sin ^2 \theta}{\cos ^{3 / 2} \theta} \times 2 \sin \theta \cos \theta d \theta \\ & =2 \times \int_0^{\pi / 2} \frac{\left(1-\cos ^2 \theta\right)}{\sqrt{\cos \theta}} \sin \theta d \theta \\ & \cos \theta=t \Rightarrow-\sin \theta d \theta=d t\end{aligned}$
$I=2 \int_1^0 \frac{t^2-1}{\sqrt{t}} d t=2\left[\frac{2}{5} t^{5 / 2}-2 \sqrt{5}\right]_1^0=\frac{16}{5}$
$I=2 \int_1^0 \frac{t^2-1}{\sqrt{t}} d t=2\left[\frac{2}{5} t^{5 / 2}-2 \sqrt{5}\right]_1^0=\frac{16}{5}$
Hence, the correct answer is option: 'D'.