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Integrals

Multiple Choice Questions

461.

$\int_{0}^{π} \frac{xdx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} dx$ is equal to

$\frac{π}{2 ab}$
$\frac{π}{ab}$
$\frac{π^{2}}{2 ab}$
$\frac{π^{2}}{ab}$

462.

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

$e^{x} \sec^{2} (\frac{x}{2}) + c$
$e^{x} \tan (\frac{x}{2})$ + c
$e^{x} \sec (\frac{x}{2}) + c$
$e^{x} \tan (x) + c$

463.

$\int \sqrt{1 + \sin (\frac{x}{4})} dx$ is equal to

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

464.
$\int_{0}^{\infty} \frac{xdx}{(1 + x) (1 + x^{2})}$ is equal to

$\frac{π}{2}$

0

1

$\frac{π}{4}$

$\frac{π}{4}$

$\int_{0}^{\frac{π}{2}} \frac{xdx}{(1 + x) (1 + x^{2})} Put x = \tan (θ) \Rightarrow \frac{dx}{dθ} = \sec^{2} (θ) = \int_{0}^{\frac{π}{2}} \frac{\tan (θ) \sec^{2} (θ) dθ}{(1 + \tan (θ)) (1 + \tan^{2} (θ))} [∵ For limit, x = 0 \Rightarrow θ = 0, x = \infty \Rightarrow θ = \frac{π}{2}] = \int_{0}^{\frac{π}{2}} \frac{\tan (θ) \sec^{2} (θ) dθ}{(1 + \tan (θ)) \sec^{2} (θ)} = \int_{}^{\frac{π}{2}} \frac{\tan (θ)}{(1 + \tan (θ))} = \int_{0}^{\frac{π}{2}} \frac{\frac{\sin (θ)}{\cos (θ)}}{1 + \frac{\sin (θ)}{\cos (θ)}} dθ = \int_{0}^{\frac{π}{2}} \frac{\sin (θ)}{\cos (θ) + \sin (θ)} dθ Let I = \int_{0}^{\frac{π}{2}} \frac{\sin (θ)}{\cos (θ) + \sin (θ)} dθ . . . (i)$

$I = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - θ)}{\cos (\frac{π}{2} - θ) + \sin (\frac{π}{2} - θ)} dθ [∵ \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] I = \int_{0}^{\frac{π}{2}} \frac{\cos (θ)}{\sin (θ) + \cos (θ)} dθ . . . (ii) Adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin (θ)}{\cos (θ) + \sin (θ)} dθ + \int_{0}^{\frac{π}{2}} \frac{\cos (θ)}{\sin (θ) + \cos (θ)} dθ = \int_{0}^{\frac{π}{2}} \frac{\sin (θ) + \cos (θ)}{\cos (θ) + \sin (θ)} dθ = \int_{0}^{\frac{π}{2}} dθ \Rightarrow 2 I = {[θ]}_{0}^{\frac{π}{2}} \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4}$