ingreate cos(logx)dx

cos(logx)dxlet logx =tthen 1xdx=dtdx=xdtdx=etdtso the integral get changed into cos(t)etdtand 1/2et(2cost)dtand 2cost can be written as {(sint+cost)+(cost-sint)}so  1/2et((sint+cost)+(cost-sint))dt (i)And et(f(t)+f'(t))=et(f(t) +CSo f(t)=sint+cost, f'(t)=cost-sintso (i) is changed into 1/2et((sint+cost)+(cost-sint))dt=(1/2)et((sint+cost))+C1/2{x(sin(logx)+cos(logx)}+C

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Bhaiya ave to h , par abhi mero btaane ko man na h .

Hopt, it is helpful to you.

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ave to hame bhi hai par thand mae hath ni chalde~!!

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hmmmmm. baat to sahi h teri

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let logx = t.; on differentiating w.r.t. x : dx/x = dt , let it be equation -1.

then , integrate cos (t) * x dt ,where x=et

=et *cost=I (say)

then apply by parts i.e.,

I=cost . et - int.(-sint.et )dt

I=cost. et + sint .et - int.(cost.et )dt which is equal to I.

2I=(cost + sint)et

I=x/2(sin(logx) + cos(logx)), where t=logx and elogx =x.

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