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: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`

If `x^2!=n pi-1, n in N`. Then, the value of `int x sqrt((2sin(x^2+1)-

1.	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: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» Correct Answer - B 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)]`

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