-2-2 e x logsin x - x dx -A- -1-2 e-4log 2B- e -4log 2C- -e-4log 2D- 1-2 e -4log 2

The correct option is D 1/2e^π/4 log 2Let I=∫_π/4^π/2 e^x (log sin x+cot x)dx I=∫_π/4^π/2 e^x log sin x dx+∫_π/4^π/2 e^x cot x dx =∫_π/4^π/2 e^x log sin xdx+[e ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

∫π/2π/2 e x l...

Question

$\int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} (l o g s i n x + c o t x) d x =$

e^{\frac{π}{4}} l o g 2

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- e^{\frac{π}{4}} l o g 2

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\frac{1}{2} e^{\frac{π}{4}} l o g 2

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- \frac{1}{2} e^{\frac{π}{4}} l o g 2

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Solution

The correct option is C $\frac{1}{2} e^{\frac{π}{4}} l o g 2$
$L e t I = \int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} (l o g s i n x + c o t x) d x$
$I = \int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} l o g s i n x d x + \int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} c o t x d x$
$= \int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} l o g s i n x d x + [e^{x} l o g s i n x]_{\frac{π}{4}}^{\frac{π}{2}}$
$- \int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} l o g s i n x d x$
$= e^{\frac{π}{2}} l o g s i n \frac{π}{2} - e^{\frac{π}{4}} l o g s i n \frac{π}{4} = \frac{1}{2} e^{\frac{π}{4}} l o g 2$

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∫π2π4 ex (log sin x+cot x)dx=

The correct option is C 12eπ4 log 2Let I=∫π2π4 ex (log sin x+cot x)dx I=∫π2π4 ex log sin x dx+∫π2π4 ex cot x dx =∫π2π4 ex log sin xdx+[ex log sin x]π2π4 −∫π2π4 ex log sin x dx =eπ2 log sin π2−eπ4 log sin π4=12eπ4log 2

$\int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} (l o g s i n x + c o t x) d x =$