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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

$\frac{1}{n}$

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

$\frac{1}{n}$

$Given, I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} (dθ) ∴ I_{n + 1} + I_{n + 1} = \int_{0}^{\frac{π}{4}} \tan^{n - 1} (θ) dθ + \int_{0}^{\frac{π}{4}} \tan^{n - 1} (θ) dθ = \int_{0}^{\frac{π}{4}} \tan^{n - 1} (θ) (1 + \tan^{2} (θ)) dθ = \int_{0}^{\frac{π}{4}} \sec^{2} (θ) \tan^{n - 1} (θ) dθ Put \tan (θ) = ϕ \Rightarrow \sec^{2} (θ) dθ = dϕ ∴ I_{n + 1} + I_{n + 1} = \int_{0}^{1} ϕ^{n - 1} dϕ = {[\frac{ϕ^{n}}{n}]}^{1}_{0} = \frac{1}{n}$