Indefinite Integration 1 Question 26

27. The value of ππcos2x1+axdx,a>0, is

(2001, 1M)

(a) π

(b) aπ

(c) π2

(d) 2π

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Solution:

  1. Let I=ππcos2x1+axdx

=ππcos2(x)1+axd(x)I=ππaxcos2x1+axdx

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

2I=ππ1+ax1+axcos2xdx=ππcos2xdx=20π1+cos2x2dx=0π(1+cos2x)dx=0π1dx+0πcos2xdx=[x]0π+20π/2cos2xdx=π+02I=πI=π/2

ecosxsinx, for |x|2



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