-cos log x d x-Fx-c- where c is an arbitrary constant- Here Fx- A- x-cos log x-sin log x-B- x-2-cos log x-sin log x-C- x-2-cos log x-sin log x-D- x-cos log x-sin log x-

The correct option is B x2 [cos (log x)+ sin (log x)]∫cos(log x)dx=I Let nbsp; log x=t ⇒ I=∫ e^tcos(t)dx= e^tcos(t)+ e^tsin(t)/2 ⇒ I=x/2 [cos (log x)+ sin (l ...

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Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

∫cos log x d ...

Question

$\int cos (log x) d x = F (x) + c,$ where $c$ is an arbitrary constant. Here $F (x) =$

x [cos (log x) + sin (log x)]

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x [cos (log x) - sin (log x)]

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\frac{x}{2} [cos (log x) + sin (log x)]

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\frac{x}{2} [cos (log x) - sin (log x)]

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Solution

The correct option is C $\frac{x}{2} [cos (log x) + sin (log x)]$
$\int cos (log x) d x = I$
Let
$log x = t \Rightarrow I = \int e^{t} cos (t) d x = \frac{e^{t} cos (t) + e^{t} sin (t)}{2} \Rightarrow I = \frac{x}{2} [cos (log x) + sin (log x)] + c$

Suggest Corrections

∫cos(logx)dx=F(x)+c, where c is an arbitrary constant. Here F(x)=

The correct option is C x2[cos(logx)+sin(logx)]∫cos(logx)dx=I Let logx=t⇒I=∫etcos(t)dx=etcos(t)+etsin(t)2⇒I=x2[cos(logx)+sin(logx)]+c

$\int cos (log x) d x = F (x) + c,$ where $c$ is an arbitrary constant. Here $F (x) =$

The correct option is C $\frac{x}{2} [cos (log x) + sin (log x)]$
$\int cos (log x) d x = I$
Let
$log x = t \Rightarrow I = \int e^{t} cos (t) d x = \frac{e^{t} cos (t) + e^{t} sin (t)}{2} \Rightarrow I = \frac{x}{2} [cos (log x) + sin (log x)] + c$