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1/(sinx + secx)

1/(sinx + secx)

Grade:12

1 Answers

Amarendra Mishra
12 Points
9 years ago
1/(sinx+secx)  =  cosx/(sinxcosx+1) = 2cosx/(2sinxcosx+2)
                                                      =(cosx +sinx)/(3-(sinx-cosx)2) + (cosx-sinx)/((cosx+sinx2)2 +1)
 
substitute sinx-cosx and t in first term and cosx-sinx as z in second term so that the integral reduces to
 
I = dt/(3-t2 ) + dz/(z2 +1)
 
then integrate using standard formulae ..
 

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