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Integrals

Multiple Choice Questions

711.

$\int \frac{x + \sin (x)}{1 + \cos (x)} dx = ?$

$xtan (\frac{x}{2}) + C$
$xsin (\frac{x}{2}) + \cos (\frac{x}{2}) + C$
$xtan (\frac{x}{2}) + \sec (\frac{x}{2}) + C$
$xtan (\frac{x}{2}) + \sec (\frac{x}{2}) + C$

712.

$If I_{n} = \int \frac{\sin (nx)}{\cos (x)} dx, then I_{n} = ?$

$\frac{- 2}{n - 1} \cos (n - 1) x - I_{n - 2}$
$\frac{2}{n - 1} \cos (n - 1) x + I_{n - 2}$
$\frac{- 2}{n + 1} \sin (n - 1) x - I_{n - 2}$
$\frac{- 2}{n + 1} \cos (n - 1) x - I_{n - 2}$

Short Answer Type

713.

$The integral \int_{0}^{2} ||x - 1| - x| d x = ?$

714.

Let [t] denote the greatest integer less than or equal to t. Then the value of $\int_{1}^{2} |2 x - [3 x]| d x$ is

Multiple Choice Questions

715.

$Area enclosed by 0 \leq y \leq x^{2} + 1, 0 \leq y \leq x + 1, \frac{1}{2} \leq x \leq 2 is :$

$\frac{1}{12}$
$\frac{1}{6}$
1
$\frac{1}{3}$

716.

$Evaluate \int_{- π}^{π} |π - |x|| d x$

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

717.

$(2^{\frac{1}{5}} + 5^{\frac{1}{8}}) has integral terms find least value of n is$

718.

$If \int \sin^{- 1} (\sqrt{\frac{x}{1 + x}}) dx = A (x) \tan^{- 1} (\sqrt{x}) + Bx + C, where C is a constant of integration, then ordered pair (A (x), B (x)) can be :$

$(x - 1, \sqrt{x})$
$(x - 1, - \sqrt{x})$
$(x + 1, \sqrt{x})$
$(x + 1, - \sqrt{x})$

719.
$If the value of the integral \int_{0}^{\frac{1}{2}} \frac{x^{2}}{{(1 - x^{2})}^{\frac{3}{2}}} dx x = \sin (θ); dx = \cos (θ) dθ$

$2 \sqrt{3} + π$

$2 \sqrt{3} - π$

$3 \sqrt{2} + π$

$3 \sqrt{2} - π$

$2 \sqrt{3} - π$

$\frac{k}{6} = \int_{0}^{\frac{1}{2}} \frac{x^{2}}{{(1 - x^{2})}^{\frac{3}{2}}} dx x = \sin (θ); dx = \cos (θ) dθ \Rightarrow \frac{k}{6} = \int_{0}^{\frac{π}{6}} \frac{\sin^{2} (θ)}{{(1 - \sin^{2} (θ))}^{\frac{3}{2}}} . \cos (θ) d θ \Rightarrow \frac{k}{6} = \int_{0}^{\frac{π}{6}} \frac{\sin^{2} (θ)}{\cos^{3} (θ)} . \cos (θ) d θ \Rightarrow \frac{k}{6} = \int_{0}^{\frac{π}{6}} \tan^{2} (θ) dθ = \int_{0}^{\frac{π}{6}} (\sec^{2} (θ) - 1) dθ \Rightarrow \frac{k}{6} = {(\tan (θ) - θ)}_{0}^{\frac{π}{6}} = (\frac{1}{\sqrt{3}} - \frac{π}{6}) = \frac{2 \sqrt{3} - π}{6} k = 2 \sqrt{3} - π$