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Integrals

Multiple Choice Questions

441.

$\int \frac{\sec (x) \csc (x)}{2 \cot (x) - \sec (x) \csc (x)} dx$ is equal to

$\log |\sec (x) + \tan (x)|$ + c
$\log |\sec (x) + \csc (x)|$ + c
$\frac{1}{2} \log |\sec (2 x) + \tan (2 x)|$ + c
$\log |\sec (2 x) + \csc (2 x)|$ + c

442.

If $\int_{0}^{a} (2 a - x) dx = μ and \int_{0}^{a} f (x) dx = λ$ , then $\int_{0}^{2 a} f (x) dx$ equals

$2 λ - μ$
$λ + μ$
$μ - λ$
$λ - 2 μ$

443.

$\int \tan (\sin^{- 1} (x)) dx$ is equal to

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

444.

$\int {(\sin (x) - \cos (x))}^{4} (\sin (x) + \cos (x)) dx$ is equal to

$\frac{\sin (x) - \cos (x)}{5} + c$
$\frac{{(\sin (x) - \cos (x))}^{5}}{5}$ + c
$\frac{{(\sin (x) - \cos (x))}^{4}}{4}$ + c
$\frac{{(\sin (x) + \cos (x))}^{5}}{5}$ + c

445.

$\int e^{\sin (θ)} [\log (\sin (θ)) + \csc^{2} (θ)] \cos (θ) dθ$ is equal to

$\int e^{\sin (θ)} [\log (\sin (θ)) + \csc^{2} (θ)] + c$
$e^{\sin (θ)} [\log (\sin (θ)) + \csc (θ)] + c$
$e^{\sin (θ)} [\log (\sin (θ)) - \csc (θ)]$
$e^{\sin (θ)} [\log (\sin (θ)) - \csc^{2} (θ)]$

446.

$\int e^{3 \log (x)} {(x^{4} + 1)}^{- 1} dx$ is equal to

e³^log(x) + c
$\frac{1}{4} \log (x^{4} + 1)$
log(x⁴ + 1) + c
$\frac{1}{2} \log (x^{4} + 1) + c$

447.
If $\int_{0}^{π} xf (\sin (x)) dx = A \int_{0}^{\frac{π}{2}} f (\sin (x)) dx$ , then A is

0

$π$

$\frac{π}{4}$

2 $π$

$π$

$Let I = \int_{0}^{π} xf (\sin (x)) dx . . . (i) = \int_{0}^{π} (π - x) f [\sin (π - x)] dx \Rightarrow I = \int_{0}^{π} (π - x) f (\sin (x)) dx . . . (ii) On adding Eqs . (i) and (ii), we get 2 I = \int_{0}^{π} π f (\sin (x)) dx$

$\Rightarrow I = \frac{π}{2} \int_{0}^{π} f (\sin (x)) dx = π \int_{0}^{\frac{π}{2}} f (\sin (x)) dx \Rightarrow A \int_{0}^{\frac{π}{2}} f (\sin (x)) dx = π \int_{0}^{\frac{π}{2}} f (\sin (x)) dx \Rightarrow A = π$