`int_(0)^(pi//2)(sinx-cosx)log(sinx+cosx)dx=0`

1.	`int_(0)^(pi//2)(sinx-cosx)log(sinx+cosx)dx=0`
Answer» Apply `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx` to get `I=-I`.

Discussion

The value of `int_0^oox/((1+x)(x^2+1))dx` isA. `2pi`B. `pi/4`C. `pi/16`D. `pi/32`
` int _(0)^(a) sqrt(a^(2) - x^(2))` dx is equal toA. `pia^(2)`B. `1/2 pia^(2)`C. `1/3 pia^(2)`D. `1/4 pia^(2)`
` int _(0)^(pi//3) (cos x + sin x)/(sqrt(1+sin 2x))dx ` is equal toA. `(4pi)/(3)`B. `(2pi)/3`C. `pi`D. `pi/3`
The value of ` int _(0)^(pi//2) ((sin x + cos x)^(2))/(sqrt(1+sin 2x) dx` is
` int _(-3pi//2)^(-pi//2) [ ( x + pi)^(3) + cos^(2) x ] ` dx is equalt toA. `((pi^(4))/32)+(pi/2)`B. ` (pi/2)`C. ` (pi/4)-1`D. ` (pi^(4))/32`
The value of ` int _(0)^(pi) (dx)/(5+4 cos x ) ` isA. `2pi`B. `(3pi)/2`C. `(5pi)/4`D. `pi/3`
If `int _(0)^(pi//2) sin^(6) dx = (5pi)/32 ,` then the value of ` int _(-pi)^(pi) (sin ^(6) x + cos^(6)x)dx` isA. `(5pi)/8`B. `(5pi)/16`C. `(5pi)/2`D. `(5pi)/4`
The value of `int_0^1 (x^4+1)/(x^2+1)dx` isA. `(1)/(6)(3pi-4)`B. `(1)/(6)(3-4pi)`C. `(1)/(6)(3pi+4)`D. `(1)/(6)(3+4pi)`
`l=int_(-2)^(1)(tan^(-1)x+"cot"^(-1)(1)/(x))dx` is equal toA. ` (5pi)/2+ 4 tan ^(-1) 2 - In 5/2`B. ` (5pi)/2 - 4 tan^(-1) 2 + In 5/2`C. ` (5pi)/2 - 3 tan ^(-1) 2 - In 5/2`D. ` (5pi)/2 - 3 tan ^(-1) 2 + 5/2`
`int_0^pi(x dx)/(a^2cos^2x+b^2sin^2x)`A. `pi/(2ab)`B. `pi/(ab)`C. `(pi^(2))/(2ab)`D. `(pi^(2))/(ab)`