IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

twinkle_star

hii plzz solve

integral of root tanx......

2) 1/ sinx+secx

3) 1/cosx+cosecx

plzz help

chimanshu_007

ans-1)put tanx = t²

=> sec²x dx = 2tdt

dx = 2tdt / 1 + t⁴

I =

t . (2t / 1 + t⁴) dt

=>

2t² / 1 + t⁴ dt

=

t² + 1 / 1 + t⁴ +

t² - 1 / 1 + t⁴ dt

=

1 + (1/t²) / t² + (1 / t²) dt +

1 - (1/t²) / t² + 1/t² dt

t² + 1/t² = (t-1/t)² + 2 = ( t + 1/t)² - 2

=

1 + (1/t²) / (t-1/t)² + 2dt +

1 - (1/t²) / ( t + 1/t)² - 2 dt

let a = t - (1/t) , b = t + (1/t)

I =

da / a² + (root2)² +

db / b² - (root2)²

= (1/2) tan^-1 (a/root 2) + 1/2root2 log (b - root2 / b + root2 ) +c

put the value of a and b 1st and then t....

ans will come...

chimanshu_007

ans3)

dx / cosx + cosecx

I =

dx / cosx + (1/sinx)

=

dx (sinx / cosx.sinx + 1)

multiply and divide by 2

=

dx (2sinx / 2 + 2 sinx.cosx)

=

dx (2sinx / 2 + sin2x)

=

{ [sinx + cosx] + [sinx - cosx]} dx / 2 + sin2x

=

(sinx + cosx / 2 + sin2x)dx +

(sinx - cosx) / 2 + sin2x) dx

=

(sinx + cosx / 3 - (1 - sin2x) ) dx +

(sinx - cosx) / (1 + ( 1+ sin2x)) dx

=

(sinx + cosx / 3 - ( sinx - cosx )² ) +

(sinx - cosx) / 1 + ( sinx + cosx)²)dx

sinx - cosx = a , sinx + cosx = b

cosx + sinx dx = da , cosx - sinx dx = db

I =

( da / 3 - a² ) -

( db / 1 + b²)

take 3 as (root3)² and solve....

ans will come....

neeraj_agarwal_1990

2)

cosx /(sinx+cosx)

now write numerator as [(cosx+sinx)+(cosx-sinx) ]/2

and break about the integral at "+".

now put sinx-cosx=t in 1st and sinx+cosx=t in 2nd integral.

and square both sides to get 2sinx.cosx =t^2-1 in 1st and 2sinx.cosx=1-t^2 in 2nd.

now substitute all values and get the answer.

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