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If g(x)=int(0)^(x)cos^(4) dt, then g(x+p...

If `g(x)=int_(0)^(x)cos^(4) dt`, then `g(x+pi)` equals

A

`g(x)+g(pi)`

B

`g(x)-g(pi)`

C

`g(x) g(pi)`

D

`g(x)//g(pi)`

Text Solution

Verified by Experts

The correct Answer is:
A
We have,
`g(x)=underset(0)overset(x)intcos^(4)`
`:. g(x+pi)=underset(0)overset(x +pi)intc os^(4)t dtunderset(0)overset(pi)int cos^(4)t dt+underset(pi)overset(x+pi)intcos^(4) t dt`
`rArr :. g(x+pi)=g(pi)+underset(0)overset(x +pi)intcos^(4)(u +pi)`, where `t=u+pi`
`rArr g(x+pi)=g(pi)+g(x)`
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OBJECTIVE RD SHARMA-DEFINITE INTEGRALS-Section I - Solved Mcqs
  1. If int(0) ^(x) f (t) dt = x + int (x ) ^(1) t f (t) dt, then the v...

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  2. Let f be a positive function. Let I1 = int(1-k)^(k) xf{x(1-x}dx, I(2...

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  3. If g(x)=int(0)^(x)cos^(4) dt, then g(x+pi) equals

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  4. If l(n)=int(0)^(pi//4) tan^(n)x dx, n in N "then" I(n+2)+I(n) equals

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  5. If rArr I(n)=int(0)^(pi//4) tan ^(n)x dx, then for any positive integ...

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  6. The value of int(-1)^(1) (d)/(dx) ("tan"^(1)(1)/(x))dx is

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  7. The vlaue of the integral int(-1)^(3) ("tan"^(1)(x)/(x^(2)+1)+"tan"^...

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  8. If =int(1)^(e) (logx)^(n) dx, "then"I(n)+nI(n-1) is equal to

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  9. If =int(0)^(1) x^(n)e^(-x)dx "for" n in N "then" I(n)-nI(n-1)=

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  10. The value of int(1//n)^((an-1)//n) (sqrt(x))/(sqrt(a-x+sqrtx))dx, is

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  11. The value of the integral int(0)^(pi//2)log |tan x cot x |dx is

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  12. If I(1)=int(x)^(1)(1)/(1+t^(2)) dt and I(2)=int(1)^(1//x) dt "for" x g...

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  13. For all values of , int(1//e)^(tanx) (t)/(1+t^(2))dt+int(1//e)^(tanx) ...

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  14. The absolute value of int(10)^(19) (cosx)/(1+x^(8))dx, is

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  15. If f(x) is an odd pefiodc function defined on the interval [T/2,T/2], ...

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  16. If int(pi//2)^(theta) sin x dx=sin 2 theta then the of theta satisfyin...

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  17. If f(x) is periodic function with period, T, then

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  18. If f(n)=(1)/(n){(n+1)(n+2)(n+3)...(n+n)}^(1//n) then lim(n to oo)f(n)...

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  19. The poins of extremum of phi(x)=int(1)^(x) e^(-t^(2)//2) (1-t^(2))dt a...

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  20. int(-2)^(2) min(x-[x],-x-[x])dx equals, where [x] represents greates i...

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