-12cos log x-x d x-A- sin log 3B- sin log 2C- cos log 3D- None of these

The correct option is B sin (log 2)Put t=log x ⇒ dt=1/xdx As x=2 ⇒ t=log 2 and x=1 ⇒ t=0, we have ∫_1^2 cos(log x)/xdx=∫_0^log 2 cos t dt=[sin t]_0^log 2=sin(l ...

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Standard XII

Mathematics

Property 1

∫12cos log x/...

Question

$\int_{1}^{2} \frac{c o s (l o g x)}{x} d x =$

sin (log 3)

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sin (log 2)

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cos (log 3)

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None of these

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Solution

The correct option is B sin (log 2)
$P u t t = l o g x \Rightarrow d t = \frac{1}{x} d x$
$A s x = 2 \Rightarrow t = l o g 2 a n d x = 1 \Rightarrow t = 0, w e h a v e$
$\int_{1}^{2} \frac{c o s (l o g x)}{x} d x = \int_{0}^{l o g 2} c o s t d t = [s i n t]_{0}^{l o g 2} = s i n (l o g 2)$

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Q. If

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then the value of k is

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∫21cos(log x)xdx=

The correct option is B sin (log 2)Put t=log x⇒dt=1xdx As x=2⇒t=log 2 and x=1⇒t=0, we have ∫21cos(log x)xdx=∫log 20 cos t dt=[sin t]log 20=sin(log 2)

$\int_{1}^{2} \frac{c o s (l o g x)}{x} d x =$

The correct option is B sin (log 2)
$P u t t = l o g x \Rightarrow d t = \frac{1}{x} d x$
$A s x = 2 \Rightarrow t = l o g 2 a n d x = 1 \Rightarrow t = 0, w e h a v e$
$\int_{1}^{2} \frac{c o s (l o g x)}{x} d x = \int_{0}^{l o g 2} c o s t d t = [s i n t]_{0}^{l o g 2} = s i n (l o g 2)$