First slide
Evaluation of definite integrals
Question

π2π2(x3+xCosx+tan5x+1)dx=

Easy
Solution

 

    π2π2(x3+xCosx+tan5x+1)dx=  0+0+0+π2π21dx=(x)π2π2=π    since x3,x cosx, tan5x are odd functiions

 

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