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Integrals

Multiple Choice Questions

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)

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

$\frac{π}{96}$

$\frac{5}{192}$

$\frac{5 π}{256}$

$\frac{5 π}{192}$

$\frac{5}{192}$

$Let I = \int_{0}^{\frac{π}{6}} \cos^{4} (3 θ) . \sin^{2} (6 θ) dθ Put 3 θ = t \Rightarrow dθ = \frac{dt}{3} ∴ I = \frac{1}{3} \int_{0}^{\frac{π}{2}} \cos^{4} (t) . \sin^{2} (2 t) dt = \frac{1}{3} \int_{0}^{\frac{π}{2}} \cos^{4} (t) . {(2 \sin (t) \cos (t))}^{2} dt = \frac{1}{3} \int_{0}^{π} \cos^{6} (t) \sin^{2} (t) dt = \frac{4}{3} [\frac{5 . 3 . 1}{8 . 6 . 4 2} \times \frac{π}{2}] [use gamma rule of integration] = \frac{5 π}{192}$