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Question

Evaluate the definite integrals.
π3π6sin x+cos xsin 2xdx


Solution

π3π6sin x+cos xsin 2xdx
We know that (sin xcos x)2=sin2x+cos2x2 sin x cos x
(sin xcos x)2=1sin 2x
sin2x=1(sin xcos x)2
Put sin x - cos x = 1
(cos x+sin x)dx=dtdx=dtcos x+sin xfor limit when x =π6t=sinπ6cosπ6=1232When,x=π3,t=sinπ3=3212I=312132sin x+cos x1t2dtcos x+sin x=312132dt1t2
(From here we can solve it in two ways)
Ist method
I=31213211t2dt=[sin1t]312132=[sin1(312)sin1{(312)}]=[sin1(312)+sin1(312)][sin1(θ)=sin1θ]=2sin1(312)
IInd method
31213211t2dt=312(312)1tt2dtAs11(t)2=11t2,therefore 11t2 is an even function. and we know that if f(x) is an even function, thenaaf(X)dx=2a0f(X)dxI=23120dt1t2=[2sin1t]3120=2sin1(312)

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