 Hello and welcome to the session. In this session we will discuss about the method for integration by substitution. A change in the variable of integration of a reduces an integral to one of the fundamental integrals. The method in which we change the variable to some other variable is called the method of substitution. Consider the integral i equal to integral fxdx. This can be transformed into another form by changing the independent variable x to t by substituting x equal to gt. This gives dx upon dt equal to g dash t that is dx is equal to g dash t dt. Now we substitute the value for x and dx in this integral. We get i is equal to integral of f of gt into g dash t dt. We generally make a substitution for a function whose derivative also occurs in the integrand. Consider the integral given by i equal to integral of cos of mxdx. Now here what we do is we take mx as t. This gives us mdx is equal to dt that is dx is equal to 1 upon m dt. Now we substitute the value for mx and dx in this integral. We get i is equal to integral of cos t into 1 upon m dt that is this is equal to 1 upon m integral cos t dt. This is equal to 1 upon m sin t plus c. Now we substitute the value for t we get i is equal to 1 upon m sin mx plus c. This c is constant of integration. So this is how we make suitable substitutions to find the value of the integrals. Next we discuss about some important integrals involving trigonometric functions and their standard integrals using substitution technique. First standard integral that we have is integral of tan x dx is equal to log modulus sec x plus c. Now let us see how do we get this. Consider integral tan x dx. This would be equal to integral sin x over cos x dx. Now we will make some substitution like here what we can do is we can substitute cos x as t. This gives us minus sin x dx is equal to dt or sin x dx is equal to minus dt. So we get integral of tan x dx is equal to minus dt over t which is equal to minus log modulus t plus c. Now we substitute the value for t equal to cos x. So this is equal to minus log modulus cos x plus c or we can also say integral of tan x dx is equal to log modulus sec x plus c. In the same way we will make some substitutions to get other standard integrals like integral of cot x dx is equal to log modulus sin x plus c. Then we have integral of sec x dx is equal to log sec x plus tan x plus c. Then next is integral of cos x dx is equal to log cos sec x minus cot x plus c. Let us try and find out the value of this i equal to integral sin x over sin x minus a dx. Firstly we put x minus a as t this gives dx equal to dt. So this i becomes equal to integral sin of a plus t over sin of t dt. This is equal to integral sin t cos a plus cos t sin a over sin t dt. This is further equal to integral cos a dt plus integral cot t sin a dt that is this is further equal to cos a integral dt plus sin a integral cot t dt. This is equal to cos a into t plus sin a into log of modulus sin t plus c. We have used the standard integral that is integral of cot x dx is equal to log modulus sin x plus c. Now we substitute the value for t as x minus a. So this becomes equal to x minus a into cos a plus sin a into log of modulus sin x minus a plus c which further is equal to x cos a minus a cos a plus sin a into log modulus sin x minus a plus c. So now the value for i becomes equal to x cos a plus sin a into log modulus sin x minus a plus c1 where the c1 is equal to c minus a cos a. Next we shall discuss integration using trigonometric identities. When the integrant involves some trigonometric functions we use some known identities to find the integral. Let's try and find out the value for this i equal to integral sin 3x into cos 4x dx. First let's recall the identity sin a into cos b is equal to half multiplied by sin of a plus b plus sin of a minus b. Now we will use this identity to solve this integral. Here we will take a as 3x and b as 4x and we substitute these values for a and b in this identity. So this becomes equal to integral of half multiplied by sin of 3x plus 4x plus sin of 3x minus 4x dx which becomes equal to half integral of sin 7x dx minus integral of sin x dx which is further equal to half multiplied by 1 upon 7 minus cos 7x plus cos x plus c which further is equal to minus 1 upon 14 cos 7x plus 1 upon 2 cos x plus c. This is the value for the integral where the c is the constant of integration. So like this we can use the trigonometric identities and solve the integrals. This completes the session. Hope you have understood the method of substitution for integration and integration using trigonometric identities.