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JH
Numerade Educator

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Problem 13 Easy Difficulty

Evaluate the integral.

$ \displaystyle \int \sin^5 t \cos^4 t\ dt $

Answer

$-\frac{1}{5} \cos ^{5} t+\frac{2}{7} \cos ^{7} t-\frac{1}{9} \cos ^{9} t+C$

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Video Transcript

let's evaluate this trick in a metric and roll by rewriting it and then using the use of So here I see that I have an odd power on the sign. So let me pull out all the factors of sign except want just one power. I need one power hanging around at the end. It's all right out here and that will come in when I using substitution. So now it looks like I can do it. You? Not yet, but it looks like we're gearing toward U equals ko society. So if I'd like to use that substitution, I should rewrite this original signed the fourth in terms of co sign. So we have sign fourth that sign square, but that and also square that whole thing. And then using the pathetic and identity you can write, this is one minus close sites where and so we can use this to rewrite sign of the fourth And then we still have co sign of the fourth year. And now we're ready to use the U substitution. So let you because I'm and then here because of the negative sign. Let's write. This is negative to you, equal scientist DT, and then we have negative one minus you square that's also squared. And then we have you to the fourth Power, do you? So let's just go ahead and multiply this zone and then we have negative in a girl. You're the fourth and then also at the end, we have you the eighth. Now let's just go ahead and use the power rule three times and then lets his push that minus sign on through minus you to the nine over nine. Plus he and then finally, the final answer will just come back to our use up and then replace you in terms of tea. So that's co signed the fifth power over five hours to co sign to the seventh power over seven and then minus co signs to the ninth Power over nine. Plus our constancy. And that's your final answer.