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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

194.
The value of the integral $\int_{0}^{1} x {(1 - x)}^{5} dx$ is equal to

$\frac{1}{6}$

$\frac{1}{7}$

$\frac{6}{7}$

$\frac{1}{42}$

$\frac{1}{42}$

$\int_{0}^{1} x {(1 - x)}^{5} dx = \int_{0}^{1} x^{(2 - 1)} . {(1 - x)}^{(6 - 1)} dx = B (2, 6) [∵ B (m, n) = \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} dx ∵ B (m, n) = \frac{\sqrt{m} \sqrt{n}}{\sqrt{m + n}}] = \frac{\sqrt{2} \sqrt{6}}{\sqrt{(2 + 6)}} = \frac{\sqrt{2} \sqrt{6}}{\sqrt{8}} = \frac{1 . 5 . 4 . 3 . 2 . 1}{7 . 6 . 5 . 4 . 3 . 2 . 1} = \frac{1}{42}$