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Let In=int tan^n x dx, (n>1). If I4+I6...

Let `I_n=int tan^n x dx, (n>1)`. If `I_4+I_6=a tan^5 x + bx^5 + C`, Where `C` is a constant of integration, then the ordered pair `(a,b)` is equal to :

A

`(-(1)/(5),1)`

B

`((1)/(5),0)`

C

`((1)/(5),-1)`

D

`(-(1)/(5),0)`

Text Solution

Verified by Experts

The correct Answer is:
B
We have, `I_(n)=int tan^(n)x dx`
`therefore I_(n)+I_(n+2)=int tan^(n)xdx + int tan^(n+2)xdx`
`" "=int tan^(n)x(1+tan^(2)x)dx`
`" "=int tan^(n) x sec^(2)xdx=(tan^(n+1))/(n+1)+C`
Put n = 4, we get `I_(4)+I_(6)=(tan^(5)x)/(5)+C`
`therefore" "a=(1)/(5)and b = 0`
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