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691.
For any integer n $\geq$ 2, let I_n = $\int$ tanⁿ(x)dx. If $I_{n} = \frac{1}{a} \tan^{n - 1} (x) - {bI}_{n - 2} for n \geq 2,$ then the ordered par (a, b) equals to

$(n - 1, \frac{n - 1}{n - 2})$

$(n - 1, \frac{n - 2}{n - 1})$

(n, 1)

(n - 1, 1)

(n - 1, 1)

$Given, I_{n} = \int \tan^{n} (x) dx = \int \tan^{n - 2} (x) . \tan^{2} (x) dx = \int \tan^{n - 2} (x) . \sec^{2} ((x) - 1) dx = \int \tan^{n - 2} (\sec^{2} (x)) dx - \int \tan^{n - 2} dx = \frac{\tan^{n - 1} (x)}{n - 1} - I_{n - 2} [∵ put \tan (x) = t \Rightarrow \sec^{2} (x) dx = dt ∴ \int \tan^{n - 2} dt = \frac{t^{n - 1}}{n - 1} = \frac{\tan {(x)}^{n - 1}}{n - 1}] But it is given I_{n} = \frac{1}{a} t {an}^{n - 1} x - {bI}_{n - 2} ∴ \frac{1}{a} = \frac{1}{n - 1} and b = 1 ∴ (a, b) = (n - 1, 1)$

692.

$If \frac{(x^{2} - 1)}{{(x + 1)}^{2} \sqrt{x (x^{2} + x + 1)}} dx = {Atan}^{- 1} (\sqrt{\frac{x^{2} + x + 1}{x}}) C, in which C is a constant, then A = ?$

$\frac{1}{2}$
3
2
1

693.

$By the defination of the definite integral, the value of \lim_{n \to \infty} (\frac{1^{4}}{1^{5} + n^{5}} + \frac{2^{4}}{2^{5} + n^{5}} + \frac{3^{4}}{3^{5} + n^{6}} + . . . + \frac{n^{4}}{n^{5} + n^{5}}) is$

$\log (2)$
$\frac{1}{5} \log (2)$
$\frac{1}{4} \log (2)$
$\frac{1}{3} \log (2)$

694.

$\int_{0}^{\frac{π}{6}} \cos^{4} (3 θ) . \sin^{2} (6 θ) dθ = ?$

$\frac{π}{96}$
$\frac{5}{192}$
$\frac{5 π}{256}$
$\frac{5 π}{192}$

695.

$\int \frac{dx}{(x - 1) \sqrt{x^{2} - 1}}$ is equal to

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

696.

$\int e^{x} \frac{x^{2} + 1}{{(x + 1)}^{2}} dx$ is equal to

$\frac{e^{x}}{x + 1} + C$
$\frac{- e^{x}}{x + 1} + C$
$e^{x} (\frac{x - 1}{x + 1}) + C$
$e^{x} (\frac{x + 1}{x - 1}) + C$

697.

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

$\log |\frac{1 + {xe}^{x}}{{xe}^{x}}| + C$
$\log |\frac{{xe}^{x}}{1 + {xe}^{x}}| + C$
$\log |{xe}^{x} (1 + {xe}^{x})| + C$
$\log |1 + {xe}^{x}| + C$

698.

$\int \frac{f (x) g' (x) - f' (x) g (x)}{f (x) g (x)} [logg (x) - \log (f (x))] dx$ is equal to

$\log [\frac{g (x)}{f (x)}] + C$
$\frac{1}{2} {[\log (\frac{g (x)}{f (x)})]}^{2} + C$
$\frac{g (x)}{f (x)} [\log (\frac{g (x)}{f (x)})] + C$
$[\log (\frac{g (x)}{f (x)})] - \frac{g (x)}{f (x)} + C$

699.

$\int_{0}^{\frac{π}{4}} \frac{\sin (x) + \cos (x)}{3 + \sin (2 x)} dx = ?$

$\frac{1}{2} \log (3)$
$\log (2)$
log(3)
$\frac{1}{4} \log (3)$

700.

$\int_{- 1}^{1} \frac{\sqrt{1 + x + x^{2}} - \sqrt{1 - x + x^{2}}}{\sqrt{1 + x + x^{2}} + \sqrt{1 - x + x^{2}}} dx = ?$

$\frac{3 π}{2}$
$\frac{π}{2}$
0
- 1

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691.
For any integer n $\geq$ 2, let I_n = $\int$ tanⁿ(x)dx. If $I_{n} = \frac{1}{a} \tan^{n - 1} (x) - {bI}_{n - 2} for n \geq 2,$ then the ordered par (a, b) equals to

$(n - 1, \frac{n - 1}{n - 2})$

$(n - 1, \frac{n - 2}{n - 1})$

(n, 1)

(n - 1, 1)

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

691. For any integer n ≥ 2, let In = ∫tann(x)dx. If In = 1atann - 1x - bIn - 2 for n≥ 2, then the ordered par (a, b) equals to n - 1, n - 1n - 2 n - 1, n - 2n - 1 (n, 1) (n - 1, 1)

691.
For any integer n $\geq$ 2, let I_n = $\int$ tanⁿ(x)dx. If $I_{n} = \frac{1}{a} \tan^{n - 1} (x) - {bI}_{n - 2} for n \geq 2,$ then the ordered par (a, b) equals to

$(n - 1, \frac{n - 1}{n - 2})$

$(n - 1, \frac{n - 2}{n - 1})$

(n, 1)

(n - 1, 1)