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Integrals

Multiple Choice Questions

921.

$If f (x) = \frac{1}{x^{2}} \int_{3}^{x} (2 t - 3 f' (t)) dt, then f' (t), then f' (3) is equal to$

$\frac{- 1}{2}$
$\frac{- 1}{3}$
$\frac{1}{2}$
$\frac{1}{3}$

922.

$\int \frac{dx}{(x + 100) \sqrt{x + 99}} = f (x) + c \Rightarrow f (x)$

2(x + 100)^1/2
3(x + 100)^1/2
$2 \tan^{- 1} (\sqrt{x + 99})$
$2 \tan^{- 1} (\sqrt{x + 100})$

923.

$\int \frac{3 - x^{2}}{1 - 2 x + x^{2}} e^{x} dx = e^{x} f (x) + c \Rightarrow f (x)$

$\frac{1 + x}{1 - x}$
$\frac{1 - x}{1 + x}$
$\frac{1 - x}{x - 1}$
$\frac{x - 1}{1 + x}$

924.

$\int \frac{\sqrt{\cot (x)}}{\sin (x) \cos (x)} dx = - f (x) + c \Rightarrow f (x)$

$2 \sqrt{\tan (x)}$
$- 2 \sqrt{\tan (x)}$
$- 2 \sqrt{\cot (x)}$
$2 \sqrt{\cot (x)}$

925.

$\int_{\frac{- π}{2}}^{\frac{π}{2}} \log (\frac{2 - \sin (θ)}{2 + \sin (θ)}) dθ$ is equal to

926.

$\int_{0}^{2} \frac{2 x - 2}{2 x - x^{2}} dx$ is equal to

927.

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

$\log |\frac{1 + \cos (x)}{\cos (x)}|$
$\log |\frac{\cos (x)}{1 + \cos (x)}|$
$\log |\frac{\sin (x)}{1 + \sin (x)}|$
$\log |\frac{1 + \sin (x)}{\sin (x)}|$

928.

$\int \frac{x^{49} \tan^{- 1} (x^{50})}{(1 + x^{100})} dx = k {(\tan^{- 1} (x^{50}))}^{2} + c$ , then k is equal to

$\frac{1}{50}$
$- \frac{1}{50}$
$\frac{1}{100}$
$- \frac{1}{100}$

929.

$\int_{0}^{\frac{π}{2}} \frac{200 \sin (x) + 100 \cos (x)}{\sin (x) + \cos (x)} dx$ is equal to

$50 π$
$25 π$
$75 π$
$150 π$

930.
$\int_{0}^{π} \frac{θsin (θ)}{1 + \cos^{2} (θ)} dθ$ is equal to

$\frac{π^{2}}{2}$

$\frac{π^{2}}{3}$

$π^{2}$

$\frac{π^{2}}{4}$

$\frac{π^{2}}{4}$

$Let I = \int_{0}^{π} \frac{θsin (θ)}{1 + \cos^{2} (θ)} dθ . . . (i) = \int_{0}^{π} \frac{(π - θ) \sin (π - θ)}{1 + \cos^{2} (π - θ)} dθ \Rightarrow I = \int_{0}^{π} \frac{(π - θ) \sin (θ)}{1 + \cos^{2} (θ)} dθ . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{π} \frac{πsin (θ)}{1 + \cos^{2} (θ)} dθ Let \cos (θ) = t \Rightarrow - \sin (θ) dθ = dt ∴ 2 I = - π \int_{- 1}^{1} \frac{1}{1 + t^{2}} dt = 2 π \int_{0}^{1} \frac{1}{1 + t^{2}} dt = 2 π {[\tan^{- 1} (t)]}_{0}^{1} = 2 π \frac{π}{4} = \frac{π^{2}}{4}$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

930. ∫0πθsinθ1 + cos2θdθ is equal to π22 π23 π2 π24

930.
$\int_{0}^{π} \frac{θsin (θ)}{1 + \cos^{2} (θ)} dθ$ is equal to

$\frac{π^{2}}{2}$

$\frac{π^{2}}{3}$

$π^{2}$

$\frac{π^{2}}{4}$