Indefinite Integration 1 Question 36

37. The value of 0π/2dx1+tan3x is

(1993, 1M)

(a) 0

(b) 1

(c) π/2

(d) π/4

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Solution:

  1. Let I=0π/211+tan3xdx=0π/211+sin3xcos3xdx

I=0π/2cos3xcos3x+sin3xdx

I=0π/2cos3π2xcos3π2x+sin3π2xdx

I=0π/2sin3xsin3x+cos3xdx

On adding Eqs. (i) and (ii), we get

2I=0π/21dx2I=[x]0π/2=π/2I=π/4



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