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Integrals

Multiple Choice Questions

651.
$\int_{0}^{\frac{π}{2}} \frac{\sin (x) - \cos (x)}{1 - \sin (x) . \cos (x)} dx$ is equal to

0

$\frac{π}{2}$

$\frac{π}{4}$

$π$

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin (x) - \cos (x)}{1 - \sin (x) . \cos (x)} dx . . . (i) On putting x = (\frac{π}{2} - x) in Eq . (i), we get I = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - x) - \cos (\frac{π}{2} - x)}{1 - \sin (\frac{π}{2} - x) . \cos (\frac{π}{2} - x)} dx = \int_{0}^{\frac{π}{2}} \frac{\cos (x) - \sin (x)}{1 - \sin (x) . \cos (x)} dx = - \int_{0}^{\frac{π}{2}} (\frac{\sin (x) - \cos (x)}{1 - \sin (x) . \cos (x)}) dx . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} 0 dx = 0 \Rightarrow I = 0$

652.

If $\int_{0}^{1} \tan^{- 1} (x) dx = p$ , then the value of $\int_{0}^{1} \tan^{- 1} (\frac{1 - x}{1 + x})$ dx is

$\frac{π}{4} + p$
$\frac{π}{4} - p$
1 + p
1 - p

653.

If f(x) = x , g(x) = sin(x), then $\int f (g (x)) dx$ is equal to

sin(x) + c
- cos(x) + c
$\frac{x^{2}}{2} + c$
x sin(x) + c

654.

The value of $\int_{0}^{\frac{π}{2}} \log (\csc (x)) d x$ is

$\frac{π}{2} \log (2)$
$πlog (2)$
$- \frac{π}{2} \log (2)$
$2 πlog (2)$

655.

$\int e^{\tan (x)} (\sec^{2} (x) + \sec^{3} (x) \sin (x)) dx$ is equal to

$\sec (x) e^{\tan (x)} + c$
$\tan (x) e^{\tan (x)} + c$
$e^{\tan (x)} + \tan (x) + c$
$(1 + \tan (x)) e^{\tan (x)} + c$

656.

$\int \frac{1}{16 x^{2} + 9} dx$ is equal to

$\frac{1}{3} \tan^{- 1} (\frac{4 x}{3}) + c$
$\frac{1}{4} \tan^{- 1} (\frac{4 x}{3}) + c$
$\frac{1}{12} \tan^{- 1} (\frac{4 x}{3}) + c$
$\frac{1}{12} \tan^{- 1} (\frac{3 x}{4}) + c$

657.

The value of $\int_{4}^{7} \frac{{(11 - x)}^{2}}{x^{2} + {(11 - x)}^{2}} dx$ is

658.

$\int (\sqrt{\tan (x)} + \sqrt{\cot (x)}) dx$

$\sqrt{2} \tan^{- 1} (\frac{\tan (x)}{\sqrt{\tan (x)}}) + C$
$\sqrt{2} \tan^{- 1} (\frac{\tan (x) - 1}{\sqrt{2 \tan (x)}}) + C$
$\frac{\tan (x)}{\sqrt{2}} . \tan^{- 1} (\frac{\cot (x) + 1}{\sqrt{2 \tan (x)}}) + C$
$\frac{\tan (x)}{\sqrt{2}} . \tan^{- 1} (\frac{\cot (x) + 1}{\sqrt{2 \tan (x)}}) + C$