 Hi and welcome to the session. I am Purva and I will help you with the following question. Prove the following. Integral limit is from 0 to pi by 4 to tan cube x dx is equal to 1 minus log 2. Now we begin with the solution. We denote left hand side by i. So we have i is equal to integral limit from 0 to pi by 4 to tan cube x dx. We can write this as this is equal to 2 integral limit from 0 to pi by 4 tan square x into tan x dx. This is further equal to 2 integral limit from 0 to pi by 4. Now we know that tan square x is equal to sex square x minus 1. So we have sex square x minus 1 into tan x dx. This is equal to 2 integral limit from 0 to pi by 4 sex square x into tan x dx minus 2 integral limit from 0 to pi by 4 tan x dx. This is further equal to. Now we denote this first integral by i1 and the second integral by i2. So we get i1 minus i2. Now we will solve i1 and i2 separately. So first we consider i1 which is equal to 2 integral limit from 0 to pi by 4 sex square x into tan x dx. Now we put tan x equal to t. So put tan x equal to t. Differentiating we get sex square x dx is equal to dt. So differentiating tan x equal to t we get sex square x dx is equal to dt. Also when x is equal to upper limit that is pi by 4 we have when x is equal to pi by 4 we have t is equal to tan pi by 4. Putting the value of x here we get t is equal to tan pi by 4 and we know that tan pi by 4 is equal to 1. So we get t is equal to 1. So when x is equal to pi by 4 we have t is equal to 1 and when x is equal to 0 we have when x is equal to 0 we have t is equal to tan 0 and we know that tan 0 is equal to 0. So we get t is equal to 0 when x is equal to 0. Putting all these values in I1 we get I1 is equal to 2 integral limit from 0 to 1 t dt and this is further equal to 2. Now integrating t we get t square upon 2 and limit is from 0 to 1. Now cancelling 2 in numerator and denominator and putting the limits we get this is equal to 1 minus 0 and this is further equal to now 1 minus 0 is 1. So we get I1 is equal to 1. Now we consider I2. So consider I2 which is equal to 2 integral limit from 0 to pi by 4 tan x dx and this is equal to 2 into. Now integral tan x is log sec x. So we get log sec x and limit is from 0 to pi by 4. Now putting the limits we get this is equal to 2 into log sec. Now upper limit is pi by 4 minus log sec and lower limit is 0. So we get 2 into log sec pi by 4 minus log sec 0 and this is further equal to 2 into log. Now sec pi by 4 is equal to root 2 minus log sec 0 is equal to 1. So we get 2 into log root 2 minus log 1 and this is further equal to 2 into log root 2 minus now log 1 is 0. So we get 0. This is equal to 2 log root 2 and we can also write this as log root 2 whole square and this is equal to log 2 because root 2 whole square is equal to 2. So we get I2 is equal to log 2. Thus I which is equal to I1 minus I2 is equal to now I1 is equal to 1. So we get 1 minus I2 is equal to log 2. So we get log 2. Thus we get I is equal to 1 minus log 2 which is equal to the right hand side. Thus we have proved that integral limit from 0 to pi by 4 to tan cube x dx is equal to 1 minus log 2. Hope you have understood the solution. Take care and God bless you.