we want to evaluate the indefinite integral or find the antiderivative when the integrand contains powers of sines and powers of cosines the trig function with the odd power indicates our strategy so notice in this example we have sine squared X so the power on sine is even and we also have cosine to the fifth X so the power on cosine is odd so to review from a previous lesson if the power of the cosine is odd we save one cosine factor and can put the remaining factors to sines the reason we'll do this is we'll let u equal sine X and therefore differential u will be equal to cosine X DX so notice that differential U is equal to cosine X DX this is the reason why we save one factor of cosine and while we're here if the power of the sine is odd then we save one sine factor and convert the remaining factors to cosines so going back to our example since we have an odd power of cosine we'll save one factor of cosine so to begin let's rewrite this as the integral of sine squared X times we'll write cosine to the fifth X as cosine to the fourth x times cosine X again the reason we're doing this is we're going to let u be equal to sine X which means differential u will be equal to cosine X DX so it's using this first equation we could replace sine squared X with u squared and then using the second equation here we could replace cosine X DX with differential U which leaves us with this cosine of the fourth X which we're going to write in terms of sine so we can then write it in terms of U so using the identity sine squared X plus cosine squared X equals 1 or more specifically cosine squared X equals 1 minus sine squared X we'll perform a substitution for or cosine to the fourth X so we'll write this as integral of sine squared X and then because cosine to the fourth X is really cosine squared x squared we can write this as the quantity 1 minus sine squared x squared and then we still have cosine X DX and now we should be right this in terms of U since u is equal to sine X this would be U squared this would be the quantity 1 minus u squared squared and this would be differential u so we'll have the integral of U squared then we have the quantity 1 minus u squared squared and then differential u so now we have some algebra here in terms of U so we'll square this binomial and then distribute the u squared so 4 1 minus u squared squared we'll have 1 minus u squared times 1 minus u squared so let's write this as the integral we have u squared then we multiply this out we'll have 1 minus u squared minus u squared that's minus 2-u squared and then negative u squared times negative u squared is plus u to the 4th differential u now we'll distribute so we'll have the integral of U to the 2nd minus 2 u to the fourth plus U to the sixth differential U and now we can finally integrate in terms of U using the power rule of integration so we'll have u to the third divided by 3 minus 2 times U to the fifth divided by 5 and then plus U to the seventh divided by 7 plus C and now let's write this in terms of X by substituting sine X for u let's write this as 1/3 sine cubed x and then minus 2/5 sine to the fifth X and then plus 1/7 sine of the seventh X plus C this would be our antiderivative so again nothing to remember here is if we have an odd factor of cosine X we save one factor of cosine X and let u equal sine X I hope you found this helpful