`int_(5)^(0)tan^(-1)xdx`

I=int_(0)^(1)tan^(-1)xdx

Let J=int_(0)^(1)cot^(-1)(1-x+x^(2))dx and K= int_(0)^(1)tan^(-1)xdx .If J=lambda K (lambda in N) , then lambda equals

STATEMENT-1 : int_(-3)^(3)|x|dx=9 STATEMENT-2 : int_(0)^(1)tan^(-1)xdx=(pi)/(4)-lnsqrt(2) STATEMENT-3 : int_(0)^(pi//2)(sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx=(pi)/(4)

If int_(0)^(1) cot^(-1)(1-x+x^(2))dx=k int_(0)^(1) tan^(-1)x dx , then k=

Evaluate : (i) int_(0)^(1)sin^(-1)xdx , (ii) int_(1)^(2)(lnx)/(x^(2))dx , (iii) int_(0)^(1)x^(2)sin^(-1)xdx .

int_(0)^(oo)((tan^(-1)x)/((1+x^(2))))dx

Evaluate: int_0^1xtan^(-1)xdx

If agt0 and A=int_(0)^(a)cos^(-1)xdx, and int_(-a)^(a)(cos^(-1)x-sin^(-1)sqrt(1-x^(2)))dx=pia-lamdaA . Then lamda is

int_(0)^( pi)x sin^(-1)xdx

int_0^1 (tan^-1x)/xdx-1/2int_0^(pi/2) t/sint dt has the value (A) -1 (B) 1 (C) 2 (D) 0