 Hello friends and how are you all today? Let us integrate the following function. Here the function to be integrated is cos raised to the power 4 2x into dx. Here we can write the function as cos square 2x raised to the power, the whole raised to the power 2 into dx right? Now we also know that cos square theta is equal to 1 plus cos 2 theta divided by 2 right? Now on using this identity we have 1 plus cos 2 theta, so it will be 4x by 2 the whole raised to the power 2 into dx. Now here we have integral of 1 plus cos 4x the whole square can be written as 1 plus 2 cos 4x plus cos 2 4x cos square 4x divided by 4 into dx. Now taking out the constant we have 1 by 4 into integral of 1 into dx plus 1 by 4 integral of 2 cos 4x into dx plus 1 by 4 integral of cos square 4x into dx. Further we have 1 by 4 now integral of 1 into dx is x plus here it will be 1 by 2 integral of cos 4x will be sin 4x by 4 isn't it? Now here we have 1 by 4 cos square x can be written as 1 plus cos 8x where x is 4x sorry theta is 4x so it will be 1 plus cos 2 theta by 2 where theta is 4x into dx. Further x by 4 plus sin 4x by 8 plus 1 by 8 into integral of 1 into dx plus 1 by 8 into integral of cos 8x dx. Further simplifying we have x by 4 plus sin 4x by 8 plus x by 8 plus now here integral of cos theta is sin theta by the derivative of theta so we have sin 8x divided by divided by 64 plus c. So on rearranging we have the answer as now here x by 4 plus x by 8 is equal to what? It is equal to 3 by 8x plus 1 by 64 sin 8x plus 1 by 8 sin 4x plus c. This is the required answer to this session. So have a good day.