IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

kghedriu

Integrate the foll fns w.r.t x

f(x)= x^2/(x sinx+ cosx)^2

f(x)= e^2x * (1+sin2x/1+cos2x)

KAB

q2) Whenever there is e^x term always try to reduce it to e^x(f(x)+f '(x))

f(x)= e^2x * (1+sin2x/1+cos2x)

Now f(x)=e^2x * (1/(1+cos2x)+sin2x/(1+cos2x))

=e^2x * ((sec²x)/2+tanx) Put e^2x *tanx=t

Then e^2x * ((sec²x)/2+tanx)dx=dt/2

So

f(x)dx =

dt/2=t/2+c=e^2x *(tanx)/2 +c

KAB

Q1)

f(x)= x^2/(x sinx+ cosx)^2

This is a tricky problem.Here we cannot go for any substitution as there is x term as well as xsinx+ cosx term.

We can write x²=(xsinx+cosx)(xsinx)-(sinx-xcosx)(xcosx)

So f(x)=(xsinx+cosx)(xsinx)-(sinx-xcosx)(xcosx)/(x sinx+ cosx)²

Now if we observe clearly f(x) is nothing but d/dx((sinx-xcosx)/(xsinx+cosx))

So

f(x)dx=(sinx-xcosx)/(xsinx+cosx)+c

aman531

awesome hit mr.kab

ank_einstein

x^2/(x sinx+ cosx)^2

= xcosx/(xsinx+cosx)^2 .secx.x

{here xcosx is the derivative of (xsinx +cosx)}

=d(xsinx+cosx)/(xsinx+cosx)^2 * (xsecx)

{integrating by parts taking xsecx as the first function}

= (xsecx) * (-1)/(xsinx+cosx) -_{[ ]}

^{[ ]} d(xsecx)/dx *(-1)/(xsinx+cosx) dx

= " " -_{[ ]}

^{[ ]} sec^2 x *(xsinx+cosx) *(-1)/(xsinx+cosx) dx

= " " -_{[ ]}

^{[ ]} -sec^2 x dx

= " " -tanx +c

was a gud one

keep posting such problems.

rate me if im correct.

ank_einstein

answer to the 2nd question:

f(x)= e^2x * (1+sin2x/1+cos2x)

=e^2x *((sinx+cosx)^2)/2.cos^2 x dx

=1/2 .e^2x *(tanx +1)^2 dx

=1/4 . e^t . (tan(t/2) +1)^2.dt {put 2x=t}

=1/4. e^2 . ( 1 + tan^t/2 +2tant/2) .dt

=1/4. e^t .( 2tant/2 + sec^2 (t/2)) dt

=1/4 . e^t .( f(t) + f'(t) ) .dt {put f(t)= 2tan(t/2)}

=1/4 .e^f(t)

=1/4. e^2tan(t/2)

=1/4. e^2tanx {since t=2x}

keep posting such gud problems.............

rate me if i m correct

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