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Integrals

Multiple Choice Questions

491.

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

$\frac{1}{2} \sin^{- 1} (\sin^{2} (x)) + C$
$\frac{1}{2} \cos^{- 1} (\sin^{2} (x)) + C$
$\tan^{- 1} (\sin^{2} (x)) + C$
$\tan^{- 1} (2 \sin^{2} (x)) + C$

492.

$\int e^{\tan^{- 1} (x)} (1 + \frac{x}{1 + x^{2}}) dx$ is equal to

${xe}^{\tan^{- 1} (x)} + c$
$e^{\tan^{- 1} (x)} + c$
$\frac{1}{2} e^{\tan^{- 1} (x)} + c$
$\frac{1}{2} {xe}^{\tan^{- 1} (x)} + c$

493.

$\int \csc (x - a) \csc (x) dx$ is equal to

$\frac{- 1}{\sin (a)} \log |\sin (x) \csc (x - a)| + c$
$\frac{- 1}{\sin (a)} \log [\sin (x - a) \sin (x)] + c$
$\frac{1}{\sin (a)} \log [\sin (x) \csc (x - a)] + c$
$\frac{1}{\sin (a)} \log [\sin (x - a) \sin (x)] + c$

494.

If f(x) = $\int_{- 1}^{x} |t| dt$ , then for any x $\geq$ 0, f(x) is equal to

1 - x²
$\frac{1}{2} (1 + x^{2})$
1 + x²
$\frac{1}{2} (1 - x^{2})$

495.
$\int_{1}^{3} \frac{\sqrt{4 - x}}{\sqrt{x} + \sqrt{4 - x}} dx$ is equal to

1

3

2

0

$Let I = \int_{1}^{3} \frac{\sqrt{4 - x}}{\sqrt{x} + \sqrt{4 - x}} dx . . . (i) \Rightarrow I = \int_{1}^{3} \frac{\sqrt{4 - (4 - x)}}{\sqrt{4 - x} + \sqrt{4 - (4 - x)}} dx [∵ \int_{a}^{b} f (x) dx = \int_{a}^{b} f (a + b - x) dx] \Rightarrow I = \int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4 - x} + \sqrt{x}} dx . . . (ii) On adding Eqs . (i) and (ii), we get \Rightarrow 2 I = \int_{1}^{3} 1 dx = {[x]}_{1}^{3} \Rightarrow I = \frac{2}{2} = 1$