I = integral of sin(nx) / (1+(pi)^x)*sinx for n=1,2,3.. then prove that summation from m=1 to 10 of I (2m+1) = 10*pi IITJEE - 2009 Multi-Option Correct Question. Thank you.

Question

SAGAR SINGH - IIT DELHI · Answer

Dear student,

In this case apply Summation of sinx series, you can learn the concept from ML Khanna as well....

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

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I = integral of sin(nx) / (1+(pi)^x)sinx for n=1,2,3.. then prove that summation from m=1 to 10 of I (2m+1) = 10pi IITJEE - 2009 Multi-Option Correct Question. Thank you.

I = integral of sin(nx) / (1+(pi)^x)sinx for n=1,2,3..

then prove that summation from m=1 to 10 of I (2m+1) = 10pi

IITJEE - 2009 Multi-Option Correct Question.

Thank you.

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I = integral of sin(nx) / (1+(pi)^x)*sinx for n=1,2,3.. then prove that summation from m=1 to 10 of I (2m+1) = 10*pi IITJEE - 2009 Multi-Option Correct Question. Thank you.

I = integral of sin(nx) / (1+(pi)^x)*sinx for n=1,2,3.. then prove that summation from m=1 to 10 of I (2m+1) = 10*pi IITJEE - 2009 Multi-Option Correct Question. Thank you.

1 Answers

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Other Related Questions on Integral Calculus

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Answer and earn gift vouchers

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Register Now & Win Upto 25% Scholorship

I = integral of sin(nx) / (1+(pi)^x)sinx for n=1,2,3.. then prove that summation from m=1 to 10 of I (2m+1) = 10pi IITJEE - 2009 Multi-Option Correct Question. Thank you.

I = integral of sin(nx) / (1+(pi)^x)sinx for n=1,2,3..

then prove that summation from m=1 to 10 of I (2m+1) = 10pi

IITJEE - 2009 Multi-Option Correct Question.

Thank you.