IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

roboashwin

Integration Of

? 1/ ( (cosx)^4 + (sinx)^4) within the limits 0 to ?/4

dutta1

roboashwin

First Question Mark Stands For INTERGRAL

Second Question Mark Stands For

flowers_rsss

icandieforu

the ans is
-pie/2root2

rhd92781

divide num and denom be cos^4x

u'll get

_{[ 0]}

^{[ /4]} sec⁴x/(tan⁴x+1) dx

=_{[ 0]}

^{[ /4]} (tan²x+1)sec²x/(tan⁴x+1) dx

Now put, tanx=t. therefore, sec²xdx=dt

Limits change frm t=0 to t=1

_{[0 ]}

^{[ 1]} (t^2+1)/(t^4+1) dt

Again divide the num and denom by t^2

_{[0 ]}

^{[ 1]} (1+1/t²)/(t²+1/t²)dt

Now write t-1/t = u

t²+1/t²=(t-1/t)²+ 2 = u²+2

therefore, (1+1/t²)dt = du

Limits change frm u=-

to u = 0

_{[- ]}

^{[ 0]} du/(u²+2)

= 1/^{[ ]}

2 * tan^-1(u/^{[ ]}

2) |_-^{0_{............(its limit - infinity to zero)}}

^₌1/^{[ ]}2 * (tan^-10 - tan^-1(-

))

=

/2

2

roboashwin

Thanks A Ton, Man :)