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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
$\frac{π}{2}$
$\frac{π}{4}$
$π$

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.

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

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

659.

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

$\frac{\sin (x) + \cos (x)}{xsin (x) + \cos (x)} + C$
$\frac{x \sin (x) - \cos (x)}{xsin (x) + \cos (x)} + C$
$\frac{\sin (x) - xcos (x)}{xsin (x) + \cos (x)} + C$
None of these

660.
$\int_{0}^{x} \frac{xdx}{1 + \cos (α) \sin (x)}, (0 < α < π)$ is equal to

$\frac{πα}{\sin (α)}$

$\frac{πα}{\cos (α)}$

$\frac{πα}{1 + \sin (α)}$

$\frac{πα}{1 - \cos (α)}$

$\frac{πα}{\sin (α)}$

$Let I = \int_{0}^{π} \frac{xdx}{1 + \cos (α) \sin (x)} . . . (i) \Rightarrow I = \int_{0}^{π} \frac{(π - x)}{1 + \cos (α) . \sin (π - x)} . . . (ii) On adding Eqs . (i) and (ii), we get ∴ 2 I = π \int_{0}^{π} \frac{dx}{1 + \cos (α) . \sin (α)} = π \int_{0}^{π} \frac{dx}{1 + \cos (α) (\frac{2 \tan (x / 2)}{1 + \tan^{2} (x / 2)})} = π \int_{0}^{π} \frac{\sec^{2} (x / 2)}{(1 + \tan^{2} (x / 2)) + \cos (α) (2 \tan (x / 2))} Put \tan (x / 2) = t \Rightarrow (1 / 2) \sec^{2} (x) dx = dt ∴ 2 I = π \int_{0}^{\infty} \frac{2 dt}{1 + t^{2} + 2 tcos (α)} I = π \int_{0}^{\infty} \frac{dt}{1 + t^{2} + 2 tcos (α)} = π \int_{0}^{\infty} \frac{dt}{{(1 + \cos (α))}^{2} + \sin^{2} (α)} = \frac{π}{\sin (α)} [\tan^{- 1} (\frac{t + \cos (α)}{\sin (α)})] = \frac{πα}{\sin (α)}$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

660. ∫0xxdx1 + cosαsinx, 0 < α < π is equal to παsinα παcosα πα1 + sinα πα1 - cosα

660.
$\int_{0}^{x} \frac{xdx}{1 + \cos (α) \sin (x)}, (0 < α < π)$ is equal to

$\frac{πα}{\sin (α)}$

$\frac{πα}{\cos (α)}$

$\frac{πα}{1 + \sin (α)}$

$\frac{πα}{1 - \cos (α)}$