int(0)^(pi/2) cos^(2) x dx...

`int_(0)^(pi/2) cos^(2) x dx`

Text Solution

Verified by Experts

`"Let I "= int_(0)^(pi//2) cos^(2) x dx`
`rArr I= int_(0)^(pi//2)cos^(2) ((pi)/(2)-x)dx`
`( :' int_(0)^(a) f(x) dx= int_(0)^(a) f(a-x)dx)`
`I = int_(0)^(pi//2) sin^(2) xdx`
Adding equations (1) and (2)
`2I =int_(0)^(pi//2) (sin^(2)x+cos^(2)x) dx`
`=int_(0)^(pi//2)1 dx " "( :' sin^(2) x+cos^(2) x=1)`
`=[x]_(0)^(pi//2) =(pi)/(2)-0 rArr I=(pi)/(4)`

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If int_(0)^(pi) x f (cos^(2) x + tan ^(4) x ) dx = k int_(0)^(pi//2) f(cos^(2) x + tan ^(4) x ) dx then k =

`pi/2`

`pi/4`

`pi`

a_(n) = int_(0)^(pi//2) (cos^(2) nx)/(sin x)dx , then a_(2)-a_(1), a_(3)-a_(2), b-(4)-a_(3) are in

A.P

G.P

H.P

none

If I_(1)=int_(0)^(pi//2) cos(sin x) dx,I_(2)=int_(0)^(pi//2) sin (cos x) dx and I_(3)=int_(0)^(pi//2) cos x dx then

`I_(1)gtI_(3)gtI_(2)`

`I_(3)gtI_(1)gtI_(2)`

`I_(1)gtI_(2)gtI_(3)`

`I_(3)gtI_(2)gtI_(1)`