Let I -0-cosx -x-2-2dx and J- -0-2sin 2x-x-1dx-then the value of -J- terms of -I- equalsJ-2IJ - 0-J-1-2-1-2-I-J-1-2-1-2-I

The correct option is C J=1/π+2+1/2 I J=2 ∫_0^π/2sin 2x/2x+2dx, put 2x =t Then J=∫_0^πsin t/t+2dt =(1/π+2 )+1/2 I ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

Let I =∫_0^πc...

Question

Let $I = \int_{0}^{π} \frac{c o s x}{(x + 2)^{2}} d x$ and $J = \int_{0}^{\frac{π}{2}} \frac{s i n 2 x}{x + 1} d x$ , then the value of 'J' terms of 'I' equals

J=2I

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J = 0

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J = \frac{1}{π + 2} + \frac{1}{2} - I

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J = \frac{1}{π + 2} - \frac{1}{2} + I

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Solution

The correct option is C $J = \frac{1}{π + 2} + \frac{1}{2} - I$

$J = 2 \int_{0}^{\frac{π}{2}} \frac{s i n 2 x}{2 x + 2} d x, p u t 2 x = t$
Then $J = \int_{0}^{π} \frac{s i n t}{t + 2} d t = (\frac{1}{π + 2}) + \frac{1}{2} - I$

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Let I=∫π0cosx(x+2)2dx and J=∫π20sin2xx+1dx, then the value of 'J' terms of 'I' equals

The correct option is C J=1π+2+12−I J=2∫π20sin2x2x+2dx,put2x=t Then J=∫π0sintt+2dt=(1π+2)+12−I

Let $I = \int_{0}^{π} \frac{c o s x}{(x + 2)^{2}} d x$ and $J = \int_{0}^{\frac{π}{2}} \frac{s i n 2 x}{x + 1} d x$ , then the value of 'J' terms of 'I' equals

The correct option is C $J = \frac{1}{π + 2} + \frac{1}{2} - I$

$J = 2 \int_{0}^{\frac{π}{2}} \frac{s i n 2 x}{2 x + 2} d x, p u t 2 x = t$
Then $J = \int_{0}^{π} \frac{s i n t}{t + 2} d t = (\frac{1}{π + 2}) + \frac{1}{2} - I$