Video: Finding the Integration of a Function Containing Reciprocal Trigonometric Functions Using Integration by Substitution

Determine ∫ (9 tan π‘₯ sec π‘₯)/(5 sec π‘₯ βˆ’ 19) dπ‘₯.

02:19

Video Transcript

Determine the integral of nine tan of π‘₯ times the sec of π‘₯ all divided by five sec of π‘₯ minus 19 with respect to π‘₯.

We’re asked to evaluate the integral of the quotient of two trigonometric functions. And we don’t know how to evaluate this integral in general. There’s a few different methods we could use. For example, we could use substitution to try and make this integral easier. However, in this case, there’s actually an easier method.

We need to notice what happens when we differentiate our denominator with respect to π‘₯. We know the derivative of the sec of π‘₯ with respect to π‘₯ is the tan of π‘₯ multiplied by the sec of π‘₯. So differentiating our denominator with respect to π‘₯, we get five tan of π‘₯ times the sec of π‘₯. And in fact, this is a scalar multiple of our numerator. This means we can use one of our integral rules to help us evaluate this integral.

We want to recall the following integral rule. The integral of 𝑓 prime of π‘₯ over 𝑓 of π‘₯ with respect to π‘₯ is equal to the natural logarithm of the absolute value of 𝑓 of π‘₯ plus a constant of integration 𝑐. And our integral was almost exactly in this form. However, instead of having the coefficient of five in our numerator, we have the coefficient of nine. However, we can fix this. We’ll start by taking the constant factor of nine outside of our integral. This gives us the following expression.

Next, we want five to be in our numerator. But that means we need to multiply our entire integral by one-fifth. This means we’ve rewritten our integral into the following expression. And if we set 𝑓 of π‘₯ to be our denominator, five times the sec of π‘₯ minus 19, then 𝑓 prime of π‘₯ is our numerator, five tan of π‘₯ times the sec of π‘₯. So we can evaluate this integral by using our integral rule.

By using our integral rule with 𝑓 of π‘₯ set to be five sec of π‘₯ minus 19, we can evaluate our integral to give us nine over five times the natural logarithm of the absolute value of five sec of π‘₯ minus 19 plus a constant of integration 𝑐. And remember, we need to multiply by nine over five because we have this factor in our integral. And we could distribute nine over five over our parentheses. However, 𝑐 is a constant of integration, so we can just add this at the end of our expression instead. And this gives us our final answer.

Therefore, we were able to show the integral of nine tan of π‘₯ times the sec of π‘₯ all over five sec of π‘₯ minus 19 with respect to π‘₯ is equal to nine over five times the natural logarithm of the absolute value of five sec of π‘₯ minus 19 plus our constant of integration 𝑐.

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