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Evaluate : `int_(0)^(pi)(x)/(a^(2)cos^(2)x+b^(*2)sin^(2)x)dx` |
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Answer» We have, `I =underset(0)overset(pi)int(x)/(a^(2)cos ^(2)x+b^(2)sin^(2)x)dx" "...(1)` `I =underset(0)overset(pi)int(pi-x)/(a^(2)cos^(2)(pi-x)+b^(2)sin^(2)(pi-x))dx` `(because underset(o)overset(a)intf(x)dt=underset(o)overset(a)intf(a-x)dx)` `I=underset(o)overset(pi)int(pi-x)/(a^(2)cos^(2)x+b^(2)sin^(2)x)dx" "...(2)` On adding equations (1) a nd (2), we get `21 =underset(0)overset(pi)int(x+pi-x)/(a^(2)cos^(2)x+b^(2)sin^(2)x)dx` Using the property, `underset(0)overset(pi)intf(x)*dx=underset(0)overset(pi)int[f(x)+(2a-x)]dx` `therefore2I=pi[underset(0)overset(pi)int(1)/(a^(2)cos^(2)x+b^(2)sin^(2)x)dx+underset(0)overset(pi//2)int(1)/(a^(2)cos^(2)(pi-x)+b^(2)sin^(2)(pi-x))dx]` `I=pi underset(0)overset(pi//2)int(dx)/(a^(2)cos^(2)x(1+(b^(2))/(a^(2))tan^(2)x))` `=(pi)/(a^(2))underset(0)overset(pi//2)int(sec^(2)x dx)/(1+(b^(2))/(a^(2))tan^(2)x)` [By putting `u=b/atanx, du =b/asec^(2)x dx` when `x=0, u=0` when `x=pi/2, u=oo` `=(pi)/(a^(2))underset(0)overset(oo)int((a)/(b)du)/(a+u^(2))` `=(pi)/(ab)underset(0)overset(oo)int(du)/(1+u^(2))=(pi)/(ab)[tan^(-1)(u)]_(0)^(oo)+c` `=(pi)/(ab)[tan^(-1),ootan^(-1)0]+c` `=(pi)/(ab)[(pi)/(2)-0]+c` `=(pi^(2))/(2ab)` |
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