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Tyra

Answered question

2021-10-14

Use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.

\int \sqrt{x} \sqrt{1 - x} d x

Answer & Explanation

Benedict

Skilled2021-10-15Added 108 answers

Step 1: Given that
$\int \sqrt{x} \sqrt{1 - x} d x$
Step 2: Solve
We have,
$\int \sqrt{x} \sqrt{1 - x} d x = \int \sqrt{x - x^{2}} d x$
$\int \sqrt{- (x^{2} - x)} d x$
$\int \sqrt{- (x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2})} d x$
$= \int \sqrt{\frac{1}{4} - {(x - \frac{1}{2})}^{2}} d x$
Substitute, 2x−1 = t, such that, 2.dx=dt
$\Rightarrow \frac{1}{2} \int \sqrt{1 - {(2 x - 1)}^{2}} d x = \frac{1}{4} \int \sqrt{1^{2} - t^{2}} d t$
$= \frac{1}{4} [\frac{t}{2} \sqrt{1 - t^{2}} + \frac{1}{2} \sin^{- 1} (t)] + C$
$= \frac{(2 x + 1)}{8} \sqrt{1 - {(2 x - 1)}^{2}} + \frac{1}{8} \sin^{- 1} (2 x - 1) + C$
$= \frac{2 x - 1}{8} \times 2 \sqrt{x - x^{2}} + \frac{1}{8} \sin^{- 1} (2 x + 1) + C$
$= \frac{(4 x - 2) \sqrt{x - x^{2}} + \sin^{- 1} (2 x - 1)}{8} + C$

Jeffrey Jordon

Expert2021-11-08Added 2605 answers

Answer is given below (on video)

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