If I 1-0-cos x - x -22 dx - I 2-0- - 2cos x sin x - x -1 d x and - I 2-2- k - Y I 1- thenA- -4B- -2C- k -1D- Y -2

The correct options are A λ=4 B μ=2 C k=1 D γ=2I_1=∫_0^πcos x(x+2)^2 dx nbsp; Put nbsp; x=2t I_1=∫_0^π/2cos 2t2(t+1)^2 dt =[ cos2t2(t+1)]_0^π/2+∫_0^π/2 2sin ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

If I 1=∫0πcos...

Question

If $I_{1} = π \int 0 \frac{cos x}{(x + 2)^{2}} d x,$ $I_{2} = π / 2 \int 0 \frac{cos x sin x}{(x + 1)} d x$ and $λ I_{2} = \frac{2}{μ + π} + k - γ I_{1},$ then

λ = 4

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μ = 2

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k = 1

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γ = 2

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Solution

The correct options are
A $λ = 4$
B $μ = 2$
C $k = 1$
D $γ = 2$
$I_{1} = π \int 0 \frac{cos x}{(x + 2)^{2}} d x$
Put $x = 2 t$
$I_{1} = π / 2 \int 0 \frac{cos 2 t}{2 (t + 1)^{2}} d t$
$= {[\frac{- cos 2 t}{2 (t + 1)}]}_{0}^{π / 2} + π / 2 \int 0 \frac{- 2 sin 2 t}{2 (t + 1)} d t$
$\Rightarrow I_{1} = \frac{1}{2 + π} + \frac{1}{2} - π / 2 \int 0 \frac{4 sin t cos t}{2 (t + 1)} d t$
$\Rightarrow 2 I_{1} = \frac{2}{2 + π} + 1 - 4 I_{2}$
$\Rightarrow 4 I_{2} = \frac{2}{2 + π} + 1 - 2 I_{1}$

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If I1=π∫0cosx(x+2)2 dx, I2=π/2∫0cosxsinx(x+1) dx and λI2=2μ+π+k−γI1, then

The correct options are A λ=4 B μ=2 C k=1 D γ=2I1=π∫0cosx(x+2)2 dx Put x=2t I1=π/2∫0cos2t2(t+1)2 dt =[−cos2t2(t+1)]π/20+π/2∫0−2sin2t2(t+1) dt ⇒I1=12+π+12−π/2∫04sintcost2(t+1) dt ⇒2I1=22+π+1−4I2 ⇒4I2=22+π+1−2I1

If $I_{1} = π \int 0 \frac{cos x}{(x + 2)^{2}} d x,$ $I_{2} = π / 2 \int 0 \frac{cos x sin x}{(x + 1)} d x$ and $λ I_{2} = \frac{2}{μ + π} + k - γ I_{1},$ then