If A is square matrix and `e^(A)` is defined as
`e^(A)=I+A+(A^(2))/(2!)+(A^(3))/(3!)+...=(1)/(2){:[(f(x),g(x)),(g(x),f(x))]:},` where
`A={:[(x,x),(x,x)]:} and 0 lt x lt 1,I` is an identity matrix.
`int (g(x)+1)sinxdx` is equal to

Options

A. `(1)/(2sqrt(e^(x)-1))-"cosec"^(-1)(e^(x))+c`
B. `(2)/(sqrt(e^(x)-e^(-x)))-"sec"^(-1)(e^(x))+c`
C. `(1)/(2sqrt(e^(2x)-1))+"sec"^(-1)(e^(x))+c`
D. none of these

Answer

C

Explanation

`A=[(x,x),(x,x)]`
`impliesA^(2)=[(2x^(2),2x^(2)),(2x^(2),2x^(2))],A^(3)=[(2^(2)x^(3),2^(2)x^(3)),(2^(2)x^(3),2^(2)x^(3))]` and so on
Then `e^(A)=I+A+(A^(2))/(2!)+(A^(3))/(3!)+ … +`
`=[(1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ... ,x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...),(x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ..., 1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...)]`
`=[((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)+(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)-(1)/(2)),((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)-(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)+(1)/(2))]`
`=(1)/(2)[(e^(2x)+1,e^(2x)-1),(e^(2x)-1,e^(2x)+1)]`
` :. f(x)=e^(2x)+1 and g(x)=e^(2x)-1`
`int(e^(2x)+1)/(sqrt(e^(2x)-1))dx=int(e^(2x))/(sqrt(e^(2x)-1))dx+(1)/(sqrt(e^(2x)-1))dx`
`=int(e^(2x))/(sqrt(e^(2x)-1))dx+int (e^(x))/(e^(x) sqrt(e^(2x)-1))dx`
`=(1)/(2sqrt(e^(2x)-1))+sec^(-1)(e^(x))+C`