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Integrals

Multiple Choice Questions

191.

$\int \sqrt{\frac{1 - x}{1 + x}} dx$ is equal to

$\sin^{- 1} (x) + \sqrt{1 - x^{2}} + C$
$\sin^{- 1} (x) - 2 \sqrt{1 - x^{2}} + C$
$2 \sin^{- 1} (x) - \sqrt{1 - x^{2}} + C$
$\sin^{- 1} (x) - \sqrt{1 - x^{2}} + C$

192.

$\int \frac{dx}{1 + \tan (x)}$ is equal to

$\frac{1}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$
$\frac{x}{2} + \frac{1}{2} \log |\cos (x) - \sin (x)| + C$
$\frac{1}{2} + \frac{1}{2} \log |\cos (x) - \sin (x)| + C$
$\frac{x}{2} + \frac{1}{2} \log |\cos (x) + \sin (x)| + C$

193.
If $\int_{0}^{a} \frac{x^{2} - 1}{1 - x} dx = - \frac{1}{2}$ , then the value of a is equal to

- 1

1

2

- 2

- 1

$\int_{0}^{a} \frac{x^{2} - 1}{1 - x} dx = - \frac{1}{2} \Rightarrow \int_{0}^{a} \frac{(x - 1) (x + 1)}{(x - 1)} dx = - \frac{1}{2} \Rightarrow \int_{0}^{a} (x + 1) dx = - \frac{1}{2} \Rightarrow {[x^{2} / 2 + x]}_{0}^{a} = - \frac{1}{2} \Rightarrow a^{2} / 2 + a = - \frac{1}{2} \Rightarrow a^{2} + 2 a + 1 = 0 \Rightarrow {(a + 1)}^{2} = 0 \Rightarrow a = - 1$