First slide
Evaluation of definite integrals
Question

The value of π/2π/2x3+xcosx+tan5x+1dx is

Moderate
Solution

Let 

 I=π/2π/2x3+xcosx+tan5x+1dx I=π/2π/2x3dx+π/2π/2xcosxdx+π/2π/2tan5xdx+π/2π/21dx

we know that,aaf(x)dx=20af(x)dx, if f(x) is even 0, if f(x) is odd 

  l=0+0+0+20π/21dxx3,xcosx and tan5(x) are odd functions 

 I=2[x]0π/2=2π2=π

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