 Hello friends and how are you all today? The question says evaluate integral one upon 3 plus 2 sin x plus cos x dx. So here let I be equal to integral 1 upon 3 plus 2 sin x plus cos x dx. Now here we put tan x by we have 1 upon 2 secant square x by 2 dx equal to d that implies the value of dx is equal to 2 dt upon 1 plus t square sin secant square x by 2 can be written as 1 plus tan square x by 2 that will be 1 plus t square. So therefore we have I equal to integral 9 place of dx we have 2 dt divided by 1 plus t square upon 3 plus 2 into sin x can be written as 2t upon 1 plus t square isn't it plus cos x can be written as 1 minus t square upon 1 plus t square. So now let us solve it further we have I equal to integral 2 dt upon 1 plus t square upon taking 1 plus t square common we have 3 into 1 plus t square plus 4t plus 1 minus t square 1 plus t square will get cancelled out and we are left with I equal to let us take out this 2 which is a constant out of the integral we have dt upon 3 plus 3t square plus 4t plus 1 minus t square we have open the bracket I is equal to 2 integral dt upon 3 plus 1 gives us 4 plus 3t square minus t square gives us 2t square plus 4t on taking out 2 common from the denominator we are left with dt upon t square plus 2t plus 2 now here if we have t plus 1 the whole square that will give us t square plus 1 plus 2t that means we will have another one remaining over here so we will add it to it and now we know that integral dt upon t square plus 1 can be written as tan inverse t plus c so here we have tan inverse here we have t that is t plus 1 plus and now substitute back the value of t we took it as tan x by 2 plus 1 plus c so the required answer to the given question is tan inverse 1 plus tan x by 2 plus c right so this completes the session hope you understood it well and enjoyed it too have a nice day ahead