-
$\log2-1$
-
$\log2$
-
$\log4-1$
-
$-\log2$
$\log2-1$
$\log2$
$\log4-1$
$-\log2$
Solution:
Let, $\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\text{x}}{(2+\sin\text{x})(1+\sin\text{x})}\text{dx}$
Let $\sin\text{x}$ then $\cos\text{x}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore the integral becomes
$\text{I}=\int\limits^1_0\frac{\text{dt}}{(2+\text{t})(1+\text{t})}$
$=\int\limits^1_0\Big[\frac{-1}{2+\text{t}}+\frac{1}{1+\text{t}}\Big]\text{dt}$
$=\big[-\log(2+\text{t})+\log(1+\text{t})\big]^1_0$
$= \big[\log(1+\text{t})-\log(2+\text{t})\big]^1_0$
$=\log2-\log3-\log1+\log2$
$=\log\frac{4}{3}$
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where $[x]$ denotes the greatest integer less than or equal to $x$. Let $f \circ:(-1,1) \rightarrow R$ be the composite function defined by $(f \circ g)(x)=f(g(x))$. Suppose $c$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is NOT continuous, and suppose $d$ is the number of points in the interval $(-1,1)$ at which $f \circ g$ is $NOT$ differentiable. Then the value of $c+d$ is. . . . .