The value of - n -110-2 n -1-2 n sin 27 x d x - n -110-2 n 2 n -1sin 27 x dx is equal toA- 0B- 27C- -54D- 54

The correct option is A 0Let I_1 =∑ ^10_n=1∫^ 2n_ 2n 1sin^27x dx, I_2 =∑ ^10_n = 1∫ ^2n+1_2nsin^27 x dx I_1 =∑ ^10_n=1∫^ 2n_ 2n 1sin^27x dx, Let x= t ⇒ dx ...

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Byju's Answer

Standard XII

Mathematics

Property 6

The value of ...

Question

The value of $10 \sum n = 1 - 2 n \int - 2 n - 1 {sin}^{27} x d x + 10 \sum n = 1 2 n + 1 \int 2 n {sin}^{27} x d x$ is equal to

27

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54

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- 54

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0

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Solution

The correct option is D $0$
Let $I_{1} = 10 \sum n = 1 - 2 n \int - 2 n - 1 {sin}^{27} x d x,$
$I_{2} = 10 \sum n = 1 2 n + 1 \int 2 n {sin}^{27} x d x$

$I_{1} = 10 \sum n = 1 - 2 n \int - 2 n - 1 {sin}^{27} x d x,$
Let $x = - t \Rightarrow d x = - d t$
$\Rightarrow I_{1} = 10 \sum n = 1 2 n \int 2 n + 1 {sin}^{27} (- t) (- d t)$
$\Rightarrow I_{1} = 10 \sum n = 1 2 n \int 2 n + 1 {sin}^{27} x d x$
$\Rightarrow I_{1} = - 10 \sum n = 1 2 n + 1 \int 2 n {sin}^{27} x d x$
$\Rightarrow I_{1} = - I_{2} \Rightarrow I_{1} + I_{2} = 0$

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The value of 10∑n=1−2n∫−2n−1sin27x dx+10∑n=12n+1∫2nsin27x dx is equal to

The correct option is D 0Let I1=10∑n=1−2n∫−2n−1sin27x dx, I2=10∑n=12n+1∫2nsin27x dx I1=10∑n=1−2n∫−2n−1sin27x dx, Let x=−t⇒dx=−dt ⇒I1=10∑n=12n∫2n+1sin27(−t)(−dt) ⇒I1=10∑n=12n∫2n+1sin27x dx ⇒I1=−10∑n=12n+1∫2nsin27x dx ⇒I1=−I2⇒I1+I2=0

The value of $10 \sum n = 1 - 2 n \int - 2 n - 1 {sin}^{27} x d x + 10 \sum n = 1 2 n + 1 \int 2 n {sin}^{27} x d x$ is equal to