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Integrals

Multiple Choice Questions

671.

$If \int (1 - \cos (x)) \csc^{2} (x) dx = f (x) + c, then f (x) is equal to$

$\tan (\frac{x}{2})$
$\cot (\frac{x}{2})$
$2 \tan (\frac{x}{2})$
$\frac{1}{2} \tan (\frac{x}{2})$

672.

$If I_{n} = \int_{0}^{\frac{n^{} π}{4}} \tan (x) dx, then I_{2} + I_{4}, I_{3} + I_{5}, I_{4} + I_{6}, . . ., are in$

arithmatic progression
geometric progression
harmonic progression
arithmetic-geometric progression

673.

$\int (\sqrt{\frac{a + x}{a - x}} + \sqrt{\frac{a - x}{a + x}}) dx = ?$

$2 \sin^{- 1} (\frac{x}{a}) + c$
$2 {asin}^{- 1} (\frac{x}{a}) + C$
$2 \cos^{- 1} (\frac{x}{a}) + C$
$2 a \cos^{- 1} (\frac{x}{a}) + C$

674.

$If \int \frac{\sin^{8} (x) - \cos^{8} (x)}{1 - 2 \sin^{2} (x) \cos^{2} (x)} dx = Asin (2 x) + B, then A = ?$

$- \frac{1}{2}$
- 1
$\frac{1}{2}$
1

675.

$\int \frac{1 + \cos (4 x)}{\cot (x) - \tan (x)} dx$

$- \frac{1}{4} \cos (4 x) + C$
$\frac{1}{8} \cos (4 x) + C$
$\frac{1}{4} \sin (4 x) + C$
$- \frac{1}{8} \cos (4 x) + C$

676.

$If I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} (dθ) for n = 1, 2, 3 . . ., then I_{n + 1} + I_{n + 1} = ?$

0
1
$\frac{1}{n}$
$\frac{1}{n + 1}$

677.

$Let f (0) = 1, f (0.5) = \frac{5}{4}, f (1) = 2, f (1.5) = \frac{13}{4}, and f (2) = 5 . Using Simson' s rule, \int_{0}^{2} f (x) dx = ?$

$\frac{14}{3}$
$\frac{7}{6}$
$\frac{14}{9}$
$\frac{7}{9}$

678.
$\int \frac{dx}{x^{2} \sqrt{4 + x^{2}}} =$

$\frac{1}{4} \sqrt{4 + x^{2}} + C$

$- \frac{1}{4} \sqrt{4 + x^{2}} + C$

$- \frac{1}{4 x} \sqrt{4 + x^{2}} + C$

$\frac{9}{4 x} \sqrt{4 + x^{2}} + C$

$- \frac{1}{4 x} \sqrt{4 + x^{2}} + C$

$Let I = \int \frac{dx}{x^{2} \sqrt{4 + x^{2}}} = \int \frac{dx}{x^{3} \sqrt{\frac{4}{X^{2}} + 1}} Put \frac{4}{X^{2}} + 1 = t \Rightarrow - \frac{8}{x^{3}} dx = dt ∴ I = \int \frac{dt}{- 8 \sqrt{t}} = - \frac{1}{8} \times \frac{\sqrt{t}}{\frac{1}{2}} + C = - \frac{1}{4} \sqrt{\frac{4}{X^{2}} + 1} + C = - \frac{1}{4 x} \sqrt{4 + x^{2}} + C$