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461.
$\int_{0}^{π} \frac{xdx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} dx$ is equal to

$\frac{π}{2 ab}$

$\frac{π}{ab}$

$\frac{π^{2}}{2 ab}$

$\frac{π^{2}}{ab}$

$\frac{π^{2}}{ab}$

$Let I = \int_{0}^{π} \frac{xdx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} dx . . . (i) I = \int_{0}^{π} \frac{(π - x)}{a^{2} \cos^{2} (π - x) + b^{2} \sin^{2} (π - x)} [Using \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] or I = \int_{0}^{π} \frac{(π - x)}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} dx . . . (ii) Adding Eqs . (i) and (ii), 2 I = \int_{0}^{π} \frac{x + π - x}{(a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x))} = π \int_{0}^{π} \frac{dx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} = 2 π \int_{0}^{\frac{π}{2}} \frac{dx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} [∵ \int_{0}^{2 a} f (x) dx = 2 \int_{0}^{a} f (x) dx,] = 2 π \int_{0}^{\frac{π}{2}} \frac{x}{a^{2} + b^{2} \tan^{2} (x)}$

$Now, put \tan x = t ∴ dx = \frac{dt}{\sec^{2} (x)} = 2 π \int_{0}^{\infty} \frac{dt}{a^{2} + b^{2} t^{2}} = \frac{1}{b^{2}} \times 2 π \int_{0}^{\infty} \frac{dt}{\frac{a^{2}}{b^{2}} + t^{2}} = [\frac{2 π}{b^{2}} . \frac{b}{a} \tan^{- 1} (\frac{bt}{a})] = \frac{2 π}{ab} [\tan^{- 1} (\infty) - \tan^{- 1} (0)] = \frac{2 π}{ab} (\frac{π}{2} - 0) = \frac{2 π}{ab} \times \frac{π}{2} = \frac{π^{2}}{ab}$

462.

$\int e^{x} (\frac{1 + \sin (x)}{1 + \cos (x)}) dx$ is equal to

$e^{x} \sec^{2} (\frac{x}{2}) + c$
$e^{x} \tan (\frac{x}{2})$ + c
$e^{x} \sec (\frac{x}{2}) + c$
$e^{x} \tan (x) + c$

463.

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

$8 (\sin (\frac{x}{8}) + \cos (\frac{x}{8})) + C$
$8 (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$
$8 (\cos (\frac{x}{8}) - \sin (\frac{x}{8})) + C$
$\frac{1}{8} (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$

464.

$\int_{0}^{\infty} \frac{xdx}{(1 + x) (1 + x^{2})}$ is equal to

$\frac{π}{2}$
0
1
$\frac{π}{4}$

465.

If I_n = $\int {(\log (x))}^{n} dx$ , then I_n + nI_{n - 1} is equal to

${(xlog (x))}^{n}$
$x {(\log (x))}^{n}$
$n {(\log (x))}^{n}$
${(\log (x))}^{n - 1}$

466.

The value of $\int \frac{dx}{\sqrt{2 x - x^{2}}}$ is

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

467.

The value of $\int xlog (x^{3}) dx$ is

$\frac{1}{16} (4 x^{4} \log (x) - x^{4} + c)$
$\frac{1}{8} (x^{4} \log (x) - 4 x^{4} + c)$
$\frac{x^{4} \log (x)}{4} + c$
$\frac{1}{16} (4 x^{4} \log (x) + x^{4} + c)$

468.

The value of $\int_{0}^{π} \frac{dx}{5 + 3 \cos (x)}$ is

$\frac{π}{4}$
$\frac{π}{8}$
$\frac{π}{2}$
zero

469.

The value of $\int_{0}^{\frac{π}{2}} \log (\tan (x)) dx$ is

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

470.

The value of $\int_{- \frac{1}{2}}^{\frac{1}{2}} \cos (x) [\log (\frac{1 - x}{1 + x})] dx$ is

2e^1/2
1
e^1/2
zero

Previous Year Papers

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Integrals

Multiple Choice Questions

461.
$\int_{0}^{π} \frac{xdx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} dx$ is equal to

$\frac{π}{2 ab}$

$\frac{π}{ab}$

$\frac{π^{2}}{2 ab}$

$\frac{π^{2}}{ab}$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

461. ∫0πxdxa2cos2x + b2sin2xdx is equal to π2ab πab π22ab π2ab

461.
$\int_{0}^{π} \frac{xdx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} dx$ is equal to

$\frac{π}{2 ab}$

$\frac{π}{ab}$

$\frac{π^{2}}{2 ab}$

$\frac{π^{2}}{ab}$