Video: Using Integration by Substitution to Evaluate the Integral of a Rational Function

Find ∫ (cos π‘₯)/(1 + sinΒ² π‘₯) dπ‘₯.

01:34

Video Transcript

Find the integral of cos of π‘₯ over one plus sin squared of π‘₯ with respect to π‘₯.

Now, this isn’t a particularly nice integral. But if we look carefully, we can see that the numerator of our fraction is equal to the derivative of part of the denominator. And this means we need to use integration by substitution to evaluate our integral. We’re going to choose our substitution to be equal to sin π‘₯. And this is because if we let 𝑒 be equal to sin π‘₯ and we differentiate 𝑒 with respect to π‘₯, we get cos of π‘₯, which is the numerator of our fraction. And then we can rewrite this slightly.

Now, d𝑒 by dπ‘₯ isn’t a fraction. But whilst doing integration by substitution, we treat it a little like one. And we can say that d𝑒 is equal to cos of π‘₯ dπ‘₯. We’re then able to replace sin of π‘₯ with 𝑒. So the denominator of our fraction becomes one plus 𝑒 squared. We can also replace cos of π‘₯ dπ‘₯ with d𝑒. And we see that we’re looking to integrate one over one plus 𝑒 squared with respect to 𝑒. Now, whilst this doesn’t look particularly nice, we can quote a general result. And that is the derivative of the inverse tan of π‘₯ is equal to one over one plus π‘₯ squared.

And so this means the antiderivative of one over one plus 𝑒 squared must be the inverse tan or the arc tan of 𝑒. And so when we integrate one over one plus 𝑒 squared, we get the arc tan or inverse tan of 𝑒 plus some constant of integration 𝑐. But remember, we defined 𝑒 to be equal to sin of π‘₯. And since we’re differentiating with respect to π‘₯, it’s common practice to leave our answer in terms of π‘₯. And so we replace 𝑒 with sin of π‘₯. And we find at the interval of cos π‘₯ over one plus sin squared π‘₯ with respect to π‘₯ is the inverse tan of sin π‘₯ plus 𝑐.

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