int_{0}^{^{n}C_{r}} { sin^{2}{x} } dx is equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The integral is $I = \int_{0}^{^{n}C_{r}} \{ \sin^{2}\{x\} \} dx$. Since $\{x\} \in [0, 1)$,$\sin^{2}\{x\} \in [0, \sin^{2} 1)$. Since $\sin^{2} 1

Vedclass · Accepted Answer

$\frac{1}{2} ^{n}C_{r}(1 - \sin 1 \cos 1)$

Vedclass · Answer

(C) The integral is $I = \int_{0}^{^{n}C_{r}} \{ \sin^{2}\{x\} \} dx$. Since $\{x\} \in [0, 1)$,$\sin^{2}\{x\} \in [0, \sin^{2} 1)$. Since $\sin^{2} 1 Thus,$I = \int_{0}^{^{n}C_{r}} \sin^{2}\{x\} dx$. Since $\sin^{2}\{x\}$ is a periodic function with period $1$,we have $I = ^{n}C_{r} \int_{0}^{1} \sin^{2} x dx$. Using the identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$,we get: $I = ^{n}C_{r} \int_{0}^{1} \frac{1 - \cos 2x}{2} dx = \frac{^{n}C_{r}}{2} [x - \frac{\sin 2x}{2}]_{0}^{1}$. $I = \frac{^{n}C_{r}}{2} (1 - \frac{\sin 2}{2}) = \frac{^{n}C_{r}}{2} (1 - \sin 1 \cos 1)$.

Column $I$	Column $II$
$(A) \int_{-1}^1 \frac{dx}{1+x^2}$	$(p) \frac{1}{2} \log \left(\frac{2}{3}\right)$
$(B) \int_0^1 \frac{dx}{\sqrt{1-x^2}}$	$(q) 2 \log \left(\frac{2}{3}\right)$
$(C) \int_2^3 \frac{dx}{1-x^2}$	$(r) \frac{\pi}{3}$
$(D) \int_1^2 \frac{dx}{x \sqrt{x^2-1}}$	$(s) \frac{\pi}{2}$

$\int_{0}^{^{n}C_{r}} \{ \sin^{2}\{x\} \} dx$ is equal to (where $\{.\}$ denotes the fractional part function and $n, r \in N$)

Similar Questions

$\int_{0}^{1/3} (\sum_{r=0}^{101} \{x + \frac{r}{3}\}) dx$ is equal to (where $\{.\}$ represents the fractional part function).

The number of continuous functions $f:[0,1] \rightarrow(-\infty, \infty)$ satisfying the condition $\int_0^1 (f(x))^2 dx = 2 \int_0^1 f(x) dx$ is

If $A_n = \int_{0}^{\pi /2} \frac{\sin((2n-1)x)}{\sin x} dx$ and $B_n = \int_{0}^{\pi /2} \left( \frac{\sin(nx)}{\sin x} \right)^2 dx$ for $n \in N$,then:

Let $f(x) = \int\limits_0^x {(t^2 + 2t + 2)dt}$ where $x$ is the set of real numbers satisfying the inequation $\log_{\sqrt{2}}(1 + \sqrt{6x - x^2 - 8}) \ge 0$. If the range of $f(x)$ is $[a, b]$,then $(a + b)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module