If I 1-0-sin 884 xsin 1122 x -sin x d x and I 2-01 x 238 x 1768-1- x 2-1 dx- then I 1- I 2 is equal toA- 1B- 2C- 3D- 4

The correct option is B 2 If I_1∫_0^πsin884x sin 1122x/sinxdx =∫_0^πcos238x/sinxdx ∫_0^πcos2006x/sinxdx Let I_2m=∫_0^πcos2mx/sinxdx then I_2m I2m 2=∫_0^πcos ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

If I 1∫0πsin ...

Question

If $I_{1} \int_{0}^{π} \frac{s i n 884 x s i n 1122 x}{s i n x} d x$ and $I_{2} \int_{0}^{1} \frac{x^{238} (x^{1768 - 1})}{x^{2} - 1} d x$ , then $\frac{I_{1}}{I_{2}}$ is equal to

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Solution

The correct option is B
2

If $I_{1} \int_{0}^{π} \frac{s i n 884 x s i n 1122 x}{s i n x} d x = \int_{0}^{π} \frac{c o s 238 x}{s i n x} d x - \int_{0}^{π} \frac{c o s 2006 x}{s i n x} d x$
Let $I_{2 m} = \int_{0}^{π} \frac{c o s 2 m x}{s i n x} d x t h e n I_{2 m} - I 2 m - 2 = \int_{0}^{π} \frac{c o s 2 m x - c o s (2 m - 2) x}{s i n x} d x = 2 {(\frac{c o s (2 m - 1) x}{2 m - 1})}_{0}^{π} = - \frac{4}{2 m - 1} ∴ I_{2006} - I_{2} = - 4 (\frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2005}) & I_{238} - I_{2} = - 4 (\frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{237}) 2 I_{1} = 4 (\frac{1}{239} + \frac{1}{241} + \dots + \frac{1}{2005})$
Now,
$I_{2} = \int_{0}^{1} (x^{238} + x^{240} + \dots + x^{2004}) d x = (\frac{1}{239} + \frac{1}{241} + \dots) ∴ \frac{I_{1}}{I_{2}} = 2$

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If I1∫π0sin884xsin1122xsinxdx and I2∫10x238(x1768−1)x2−1dx, then I1I2 is equal to

If $I_{1} \int_{0}^{π} \frac{s i n 884 x s i n 1122 x}{s i n x} d x$ and $I_{2} \int_{0}^{1} \frac{x^{238} (x^{1768 - 1})}{x^{2} - 1} d x$ , then $\frac{I_{1}}{I_{2}}$ is equal to