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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

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

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

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

$We have, \int e^{x} (\frac{1 + \sin (x)}{1 + \cos (x)}) dx = \int e^{x} (\frac{(1 + 2 \sin (\frac{x}{2}) \cos (\frac{x}{2}))}{2 \cos^{2} (\frac{x}{2})}) [∵ \sin (2 x) = 2 \sin (x) \cos (x), 1 + \cos (2 x) = 2 \cos^{2} (x)] = \int e^{x} [(\frac{1}{2 \cos^{2} (\frac{x}{2})}) + (\frac{2 \sin (\frac{x}{2}) \cos (\frac{x}{2})}{2 \cos^{2} (\frac{x}{2})})] dx = \int e^{x} ((\frac{1}{2}) \sec^{2} (\frac{x}{2}) + \tan (\frac{x}{2})) ∴ e^{x} \{f (x) + f' (x)\}, dx = e^{x} f (x) + c Here, f (x) = \tan (\frac{x}{2}) and f' (x) = \sec^{2} (\frac{x}{2}) \Rightarrow e^{x} (\tan (\frac{x}{2})) + C ∴ \int e^{x} (\frac{1 + \sin (x)}{1 + \cos (x)}) dx = e^{x} \tan (\frac{x}{2}) + c$