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Integrals

Multiple Choice Questions

651.

$\int_{0}^{\frac{π}{2}} \frac{dx}{1 + \tan^{3} (x)} = ?$

$π$
$\frac{π}{2}$
$\frac{π}{4}$
$\frac{3 π}{2}$

652.

$\int_{- 1}^{1} \frac{\cos (hx)}{1 + e^{2 x}} d x = ?$

0
1
$\frac{e^{2} - 1}{2 e}$
$\frac{e^{2} + 2}{2 e}$

653.
$If \int \frac{e^{x} - 1}{e^{x} + 1} dx = f (x) + c, then f (x) is equal to$

$2 \log (e^{x} - 1)$

$2 \log (e^{2 x} - 1)$

$2 \log (e^{x} + 1) - x$

$2 \log (e^{2 x} + 1)$

$2 \log (e^{x} + 1) - x$

$We have, \int \frac{e^{x} - 1}{e^{x} + 1} dx = \int (\frac{2 e^{x}}{e^{x} + 1} - 1) dx - x + c On compairing with f (x) + c, we get f (x) = 2 \log (e^{x} + 1) - x$

654.

$\int \tan^{- 1} (\sqrt{\frac{1 - x}{1 + x}}) dx = ?$

$\frac{1}{2} ({xcos}^{- 1} (x) - \sqrt{1 - x^{2}}) + c$
$\frac{1}{2} ({xcos}^{- 1} (x) + \sqrt{1 - x^{2}}) + c$
$\frac{1}{2} (x \sin^{- 1} (x) - \sqrt{1 - x^{2}}) + c$
$\frac{1}{2} (x \sin^{- 1} (x) + \sqrt{1 - x^{2}}) + c$

655.

$\frac{\sin (x) + 8 \cos (x)}{4 \sin (x) + 6 \cos (x)} dx = ?$

$x + \frac{1}{2} \log (4 \sin (x) + 6 \cos (x)) + c$
$2 x + \log (2 \sin (x) + 3 \cos (x)) + c$
$x + \log (2 \sin (x) + 3 \cos (x)) + c$
$\frac{1}{2} \log (4 \sin (x) + 6 \cos (x)) + c$

656.

$If f (t) = \int_{- t}^{t} \frac{e^{- |x|}}{2} dx, then \lim_{t \to \infty} f (t) = ?$

1
$\frac{1}{2}$
0
- 1

657.

$\int_{0}^{2 π} \sin^{6} (x) \cos^{5} (x) dx = ?$

$2 π$
$\frac{π}{2}$
0
$- π$

658.

If $\int e^{x} (\frac{1 - \sin (x)}{1 - \cos (x)}) dx = f (x) + constant$ , then f(x) is equal to

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

659.

If I_n = $\int x^{n} e^{cx} dx$ for n $\geq$ 1, then cI_n + n . I_n_{- 1} is equal to

xⁿe^cx
xⁿ
e^cx
xⁿ + e^cx

660.

If $\int e^{x} (1 + x) . \sec^{2} ({xe}^{x}) dx$ = f(x) + constant, then f(x) is equal to

$\cos ({xe}^{x})$
$\sin ({xe}^{x})$
$2 \tan^{- 1} (x)$
$\tan ({xe}^{x})$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

653. If ∫ex - 1ex + 1dx = fx + c, then fx is equal to 2logex - 1 2loge2x - 1 2logex + 1 - x 2loge2x + 1

653.
$If \int \frac{e^{x} - 1}{e^{x} + 1} dx = f (x) + c, then f (x) is equal to$

$2 \log (e^{x} - 1)$

$2 \log (e^{2 x} - 1)$

$2 \log (e^{x} + 1) - x$

$2 \log (e^{2 x} + 1)$