Question
$\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$ is equal to:
-
$\frac{\pi}{4}$
-
$\frac{\pi}{2}$
-
$\pi$
-
$1$
$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\pi$
$1$
Solution:
We have,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_01(-\cos\text{x})\text{dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\int\limits^{\frac{\pi}{2}}_0\cos\text{x dx}$
$=-\big[\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\big[\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$=-\big[0-0\big]+\big[1-0\big]$
$=1$
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$f_n(x)=\sum_{j=1}^n \tan ^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right) \text { for all } x \in(0, \infty)$
(Here, the inverse trigonometric function $\tan ^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ )
Then, which of the following statement(s) is (are) TRUE?
$(A)$ $\sum_{ j =1}^5 \tan ^2\left( f _{ j }(0)\right)=55$
$(B)$ $\sum_{ j =1}^{10}\left(1+ f _{ j }^{\prime}(0)\right) \sec ^2\left( f _{ j }(0)\right)=10$
$(C)$ For any fixed positive integer $n$, $\lim _{x \rightarrow \infty} \tan \left(f_n(x)\right)=\frac{1}{n}$
$(D)$ For any fixed positive integer $n, \lim _{x \rightarrow \infty} \sec ^2\left(f_n(x)\right)=1$
$\frac{\pi}{3}$
$\frac{\pi}{6}$
$\frac{\pi}{12}$
$\frac{\pi}{2}$