 Hi, and how are you all today? My name is Priyanka and let us discuss this question. It says evaluate integral sine 2x upon a plus b cos x the whole square dx. Here we are given i as integral sine 2x upon a plus b cos x the whole square dx. This implies is i is equal to, now in place of sine 2x we can write 2 sine x cos x right? divided by a plus b cos x the whole square dx. Further let us put a plus b cos x equal to t. So we have minus b sine x dx equal to dt also we have the value of cos x as t minus a upon b right? So on substituting we have it as i is equal to 2 integral t minus a upon b t upon t square. So we have i is equal to 2 upon b integral t minus a upon t square dt which can be written as i is equal to 2 upon b sorry it had minus also minus 2 upon b on separating the integral sine we have t upon t square dt minus integral a upon t square dt which further implies i is equal to minus 2 upon b the bracket integral 1 upon t dt minus a integral 1 upon t dt. i is equal to minus 2 upon b square the bracket integral 1 upon t dt is log t minus a into t raise to the power minus 1 upon minus 1 plus c. i is equal to minus 2 upon b square log t is equal to log in the place of t now we have a plus b cos x minus a upon so it will be plus over here now minus minus will become plus a upon a plus b cos x plus c. So this is the required answer to the given question right so hope you understood it well have a nice day