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Integrals

Multiple Choice Questions

951.

$\int_{0}^{π} \frac{1}{1 + \sin (x)} d x = ?$

952.

$The line x = \frac{π}{4} divides the area of the region bounded by y = \sin (x), y = \cos (x) and x - axis (0 \leq x \frac{π}{2}) into two regions of areas A_{1} and A_{2} . Then A_{1}, A_{2} equals$

4 : 1
3 : 1
2 : 1
1 : 1

953.

The velocity of a particle which starts from rest is given by the following table

t (in sec) 0 2 4 6 8 10

v (in m/s) 0 12 16 20 35 60

The total distance travelled (in metre) by the particles in 10s, using trapezoidal rule is given by

954.

$If \int \frac{7 x^{8} + 8 x^{7}}{{(1 + x + x^{8})}^{2}} dx = f (x) + c, then f (x) = ?$

$\frac{x^{8}}{1 + x + x^{8}}$
$28 \log (1 + x + x^{8})$
$\frac{1}{1 + x + x^{8}}$
$\frac{- 1}{1 + x + x^{8}}$

955.

$If f_{n} (x) = \log \log \log . . . \log (x) (\log is repeated n - times), then \int {({xf}_{1} (x) f_{2} (x) . . . f_{n} (x))}^{- 1} dx is equal to$

$f_{n + 1} + c$
$\frac{f_{n + 1} (x)}{n + 1} + c$
${nf}_{n (x)} + c$
$\frac{f_{n} (x)}{n} + c$

956.

$If \int (1 - \cos (x)) \csc^{2} (x) dx = f (x) + c, then f (x) is equal to$

$\tan (\frac{x}{2})$
$\cot (\frac{x}{2})$
$2 \tan (\frac{x}{2})$
$\frac{1}{2} \tan (\frac{x}{2})$

957.
$If I_{n} = \int_{0}^{\frac{n^{} π}{4}} \tan (x) dx, then I_{2} + I_{4}, I_{3} + I_{5}, I_{4} + I_{6}, . . ., are in$

arithmatic progression

geometric progression

harmonic progression

arithmetic-geometric progression

harmonic progression

$I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} (x) dx We have, I_{r + 2} = \int_{0}^{\frac{π}{4}} \tan^{r + 2} (x) dx = \int_{0}^{\frac{π}{4}} \tan^{r} (x) . \tan^{2} (x) and I_{r} = \int_{0}^{\frac{π}{4}} \tan^{r} (x) dx Then, I_{r} + I_{r + 2} = \int_{0}^{\frac{π}{4}} \tan^{r} (x) dx + \int_{0}^{\frac{π}{4}} \tan^{r} (x) . \tan^{2} (x) = \int_{0}^{\frac{π}{4}} \tan^{r} (x) (1 + \tan^{2} (x)) dx = \int_{0}^{\frac{π}{4}} \tan^{r} (x) \sec^{2} (x) dx [Put t = \tan (x) \Rightarrow dt = \sec^{2} (x) dx] = \int_{0}^{1} t^{r} dt \Rightarrow {[\frac{t^{r + 1}}{r + 1}]}^{1}_{0} = \frac{1}{r + 1} So, I_{r} + I_{r + 2} = \frac{1}{r + 1} ie, I_{2} + I_{4} = \frac{1}{3} I_{3} + I_{5} = \frac{1}{4} I_{4} + I_{6} = \frac{1}{5} which are clearly in HP .$