Here are two problems that are very similar. We're going to go zero to π thirds of sine squared X / sqrt 1 plus cosine XDX. If we multiply by the conjugate of the denominator sqrt 1 minus cosine X on top and bottom, we can see that the denominator is going to turn into sqrt 1 minus cosine squared X, which is going to be the square root of sine squared X, which is really just sine X. So one of the sine XS in the denominator is going to cancel with one of the sine XS in the numerator, leaving us just zero to π thirds of sine X square root 1 minus cosine X DX. If we let U equal 1 minus cosine X, we can then see DU as sine X DX. Changing our bounds. We know that one minus cosine of 0 is going to be 01, minus cosine of π thirds is going to be 1/2. So we get square root of UDU. When we integrate that, we get 2/3 U to the three halves from from zero to 1/2. So now we're going to put in the upper bound minus the lower bound 1/2 to the three halves is just 1/2, sqrt 1/2. And then when we have 1/2 sqrt 1/2, we're going to rationalize that to get the root 2 / 2. And so it's going to end up being root 2 / 6, another one that's very similar zero to π halves sine squared XDX over sqrt 1 minus cosine X, multiplying by its conjugate, which is one plus cosine X underneath the square root. The denominator once again turns into plain old sine X with enough substitution, and one of those will cancel with the numerator, leaving a sine X sqrt 1 plus cosine XDX. So if we let U equal 1 plus cosine X, then DU is negative sine XDX. Changing our bounds, we get 2 to one. So when we substitute we get negative bottom bound of two up to one square root of UDU. When we take the integration, we get -2 thirds U to the three halves from 2:00 to 1:00. Plugging that in and multiplying it through, we get 4 root 2 -, 2 / 3.