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Integrals

Multiple Choice Questions

821.

$\int_{0}^{\frac{π}{2}} \frac{\tan^{7} (x)}{\cot^{7} (x) + \tan^{7} (x)} dx$ is equal to

$\frac{π}{4}$
$\frac{π}{2}$
$\frac{π}{6}$
$\frac{π}{3}$

822.

$\int {xsec}^{2} (x) dx$ is equal to

$xtan (x) + \log (\sec (x)) + c$
$\frac{x^{2}}{2} \sec (x) + \log (\cos (x)) + c$
$xtan (x) + \log (\cos (x)) + c$
$\tan (x) + \log (\cos (x)) + c$

823.

$\int \frac{t}{e^{3 t^{2}}} dt$ is equal to

$\frac{1}{6} e^{3 t^{2}} + c$
$- \frac{1}{6} e^{3 t^{2}} + c$
$\frac{1}{6} e^{- 3 t^{2}} + c$
$- \frac{1}{6} e^{- 3 t^{2}} + c$

824.

$\int_{0}^{π} \log (\sin^{2} (x)) dx$ is equal to

$2 {πlog}_{e} (\frac{1}{2})$
${πlog}_{e} (2)$
$\frac{π}{2} \log_{e} (\frac{1}{2})$
None of these

825.

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

$\frac{1}{n} \log (\frac{x^{n}}{x^{n} + 1}) + c$
$\frac{1}{n} \log (\frac{x^{n} + 1}{x^{n}}) + c$
$\log (\frac{x^{n}}{x^{n} + 1}) + c$
None of these

826.

$\int_{0}^{2} [x^{2}] dx$ is equal to

$2 - \sqrt{2}$
$2 + \sqrt{2}$
$\sqrt{2} - 1$
$- \sqrt{2} - \sqrt{3} + 5$

827.
In = $\int_{0}^{\frac{π}{4}} \tan^{n} (x) dx$ , then $\lim_{n \to \infty} n [I_{n} + I_{n + 2}]$ equals

1/ 2

2 sq units

3 sq units

4 sq uits

4 sq uits

$Given, I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} (x) dx ∴ I_{n + 2} = \int_{0}^{\frac{π}{4}} \tan^{n + 2} (x) dx = \int_{0}^{\frac{π}{4}} \tan^{n} (x) \tan^{2} (x) dx = \int_{0}^{\frac{π}{4}} \tan^{n} (x) dx (\sec^{2} (x) - 1) dx = \int_{0}^{\frac{π}{4}} \tan^{n} (x) \sec^{2} (x) dx - \int_{0}^{\frac{π}{4}} \tan^{n} (x) dx = \int_{0}^{\frac{π}{4}} \tan^{n} (x) \sec^{2} (x) dx - I_{n} \Rightarrow I_{n} + I_{n + 2} = {[\frac{\tan^{n + 1} (x)}{n + 1}]}_{0}^{\frac{π}{4}} = \frac{1}{n + 1} \lim_{n \to \infty} n [I_{n} + I_{n + 2}] = \lim_{n \to \infty} \frac{n}{n + 1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = 1$