If `x^2!=n pi-1, n in N`. Then, the value of `int x sqrt((2sin(x^2+1)-sin2(x^2+1))/(2sin(x^2+1)+sin2(x^2+1)))dx` is equal to:

Options

A. `(1)/(2)log_(e)|sec(x^(2)-1)|+C`
B. `log_(e)|sec((x^(2)-1)/(2))|+C`
C. `log_(e)|(1)/(2)sec^(2)(x^(2)-1)|+C`
D. `(1)/(2)log_(e)|sec^(2)((x^(2)-1)/(2))|+C`

Answer

B

Explanation

Let `I = intx sqrt((2 sin(x^(2)-1)-sin 2(x^(2)-1))/(2 sin(x^(2)-1)+sin 2(x^(2)-1)))dx`
Put `(x^(2)-1)/(2)=theta rArr x^(2)-1 = 2theta rArr 2x dx = 2 d theta`
`rArr" ""x dx"=d theta`
Now, `I = intsqrt((2sin 2 theta - sin 4 theta)/(2 sin 2 theta + sin 4 theta))d theta`
`=int sqrt((2sin 2 theta - 2 sin 2 theta cos 2 theta)/(2sin 2 theta + 2 sin 2 theta cos 2 theta))d theta" "(therefore sin 2 A = 2 sin A cos A)`
`=int sqrt((2 sin 2 theta(1-cos 2 theta))/(2 sin 2 theta(1+cos 2 theta)))d theta`
`=int sqrt((1-cos 2 theta)/(1+cos 2 theta))d theta = int sqrt((2sin^(2)theta)/(2 cos^(2)theta))d theta`
`[therefore 1-cos 2 A = 2 sin^(2)A and 1 + cos 2 A = 2 cos^(2) A]`
`=int sqrt(tan^(2)theta) d theta = int tan theta d theta`
`=log_(e)|sec theta|+C=log_(e)|sec((x^(2)-1)/(2))|+C[therefore theta = (x^(2)-1)/(2)]`