Indefinite Integration 1 Question 50

51. The option(s) with the values of a and L that satisfy the equation 04πet(sin6at+cos4at)dt0πet(sin6at+cos4at)dt=L, is/are

(2015 Adv.)

(a) a=2,L=e4π1eπ1

(b) a=2,L=e4π+1eπ+1

(c) a=4,L=e4π1eπ1

(d) a=4,L=e4π+1eπ+1

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Solution:

  1. Let I1=04πet(sin6at+cos4at)dt

=0πet(sin6at+cos4at)dt

+π2πet(sin6at+cos4at)dt+2π3πet(sin6at+cos4at)dt+3π4πet(sin6at+cos4at)dt

I1=I2+I3+I4+I5

Now, I3=π2πet(sin6at+cos4at)dt

Put t=π+xdt=dx

I3=0πeπ+x(sin6at+cos4at)dt=eπI2 (ii)

Now, I4=2π3πet(sin6at+cos4at)dt

Put t=2π+xdt=dx

I4=0πex+2π(sin6at+cos4at)dt=e2πI2

and I5=3π4πet(sin6at+cos4at)dt

Put t=3π+x

I5=0πe3π+x(sin6at+cos4at)dt=e3πI2 From Eqs. (i), (ii), (iii) and (iv), we get

I1=I2+eπI2+e2πI2+e3πI2=(1+eπ+e2π+e3π)I2L=04πet(sin6at+cos4at)dt0πet(sin6at+cos4at)dt=(1+eπ+e2π+e3π)=1(e4π1)eπ1 for aR



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