Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Integrals

Multiple Choice Questions

691.

The value of $\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\sin^{- 1} (x)}{{(1 - x^{2})}^{\frac{3}{2}}} dx$ is

$\frac{π}{2} - \log (2)$
$\frac{π}{4} - \frac{1}{2} \log (2)$
$\frac{π}{4} + \frac{1}{2} \log (2)$
$π - \frac{1}{2} \log (2)$

692.

Integral of $\frac{1}{2 + \cos (x)}$

$\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{1}{2} \tan (x)) + C$
$\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}} \tan (\frac{x}{2})) + C$
$- \sin (x) \log (2 + \cos (x)) + C$
$\sin (x) \log (2 + \cos (x)) + C$

693.

$\int \frac{dx}{\sqrt{x^{4} + x^{6}}}$ is equal to

$- \frac{\sqrt{1 + x^{2}}}{x} + C$
$\frac{\sqrt{1 + x^{2}}}{x} + C$
$- \frac{\sqrt{1 - x^{2}}}{x} + C$
$- \frac{\sqrt{x^{2} - 1}}{x} + C$

694.

If $\int \sin^{- 1} (x) \cos^{- 1} (x) dx = f^{- 1} (x) [\frac{π}{2} x - x f^{- 1} (x) - 2 \sqrt{1 - x^{2}}]$

$\frac{π}{2} \sqrt{1 - x^{2}} + 2 x + C$ , then

f(x) = sin(x)
f(x) = cos(x)
f(x) = tan(x)
None of these

695.

If $\int_{0}^{π} xf (\sin^{2} (x) + \sec^{2} (x)) dx = k \int_{0}^{\frac{π}{2}} f (\sin^{2} (x) + \sec^{2} (x)) dx$ , then the value of k is

$\frac{π}{2}$
$π$
$- \frac{π}{2}$
None of these

696.

$\int \frac{x + 2}{(x^{2} + 3 x + 3) \sqrt{x + 1}} dx$ is equa to

$\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x}{x + 1}) + C$
$\frac{2}{\sqrt{3}} \tan^{- 1} [\frac{x}{\sqrt{3 (x + 1)}}] + C$
$\frac{2}{\sqrt{3}} \tan^{- 1} [\frac{x}{{(x + 1)}^{2}}] + C$
None of these

697.

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

$[\tan^{- 1} (\sqrt{\frac{x}{a}} + \sqrt{\frac{x}{a}})] + C$
$a [\tan^{- 1} (\sqrt{\frac{x}{a}} - \sqrt{\frac{x}{a}})] + C$
$a [\tan^{- 1} (\sqrt{\frac{x}{a}} . \frac{(a + x)}{a})] + C$
$a [\tan^{- 1} (\sqrt{\frac{x}{a}} . \frac{(a + x)}{a} - \sqrt{\frac{x}{a}})] + C$

698.

$\lim_{n \to \infty} \frac{1}{n} (\frac{1}{n + 1} + \frac{2}{n + 2} + . . . + \frac{3 n}{4 n})$ is equal to

log(4)
- log(4)
1 - log(4)
None of these

699.
The value of the integral $\int_{0}^{\frac{π}{2}} \frac{\sin^{2} (x)}{\sin (x) + \cos (x)} dx$ is equal to

$\sqrt{2} (\log (\sqrt{2}))$

$\sqrt{2} (\sqrt{2} + 1)$

$\log (\sqrt{2} + 1)$

None of the above

None of the above

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (x)}{\sin (x) + \cos (x)} dx . . . (i) Now, I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} dx \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} (x)}{\sin (x) + \cos (x)} dx . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (x) + \cos^{2} (x)}{\sin (x) + \cos (x)} dx = \int_{0}^{\frac{π}{2}} \frac{1}{\sin (x) + \cos (x)} dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{1}{\cos (x - \frac{π}{4})} dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \sec (x - \frac{π}{4}) dx = \frac{1}{\sqrt{2}} {[\log \{\sec (x - \frac{π}{4}) + \tan (x - \frac{\log () π}{4})\}]}_{0}^{\frac{π}{2}} = \frac{1}{\sqrt{2}} [\log (\sqrt{2} + 1) - \log (\sqrt{2} - 1)] = \frac{1}{\sqrt{2}} (\log (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})) = \frac{1}{\sqrt{2}} \log {(\sqrt{2} + 1)}^{2} = \sqrt{2} \log (\sqrt{2} + 1)$