 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says integrate the following function that is sin inverse 2x divided by 1 plus x square. So let us start with the solution to this question. Let I be equal to integral of sin inverse of 2x by 1 plus x square dx. First of all we put x equal to tan t. So we will have dx will be equal to secant square t dt. We have differentiated both the sides. Therefore I becomes integral sin inverse of we put x equal to tan t. So we have 2 tan t divided by 1 plus tan square t into secant square t dt. This can be further written as integral of sin inverse into. Now this is a formula for sin 2 t. Sin 2 t is equal to 2 tan t divided by 1 plus tan square t. Now secant square t dt remains as it is. That is equal to integral of now sin inverse gets cancelled with sin we have 2t secant square t dt. 2 being a constant comes out of the integral sin we have integral of t into secant square t dt. Now by integration by parts we see that t will be the first function and secant square t will be the second function because algebraic function is given preference over trigonometric function. So by integration by parts we have 2 into first function that is t into integral of secant square t dt minus integral of dy dt of t into integral secant square t dt the whole into dt. Now this will be equal to 2 into t into integral of secant square t dt is tan t minus integral of 1 into integral of secant square t dt we have just seen is tan t dt. This is equal to 2 into t tan t plus 2 into log of mod cos t plus c because integral of tan t dt is minus log minus log of mod cos t. So this can be further written as when opening the brackets we see it is 2t into tan t plus 2 log of mod cos t plus c. Now putting back the value of t in the terms of x we will have 2 tan inverse x because t was equal to tan inverse x into x plus 2 log mod 1 upon 1 plus x square plus c. Now let us see how do we get this because cos t is equal to 1 upon secant t secant t in the denominator can be written as square root of 1 plus tan square t. Now tan t was equal to x so we have 1 upon square root of 1 plus x square. So this is how we get this this is equal to 2 x tan inverse x plus 2 log of 1 plus x square the whole raise to power minus 1 by 2 plus the constant c this is further written as 2 x into tan inverse x plus 2 into minus 1 by 2 log of 1 plus x square plus c and this can be further written as 2 gets cancelled with 2 we have 2 x tan inverse x minus log of 1 plus x square plus c. So this is our answer to the question I hope that you understood the question and enjoyed the session have a good day.