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

Determine ∫ (βˆ’7 secΒ² 3π‘₯)/(5 tan 3π‘₯) dπ‘₯.

03:10

Video Transcript

Determine the indefinite integral of negative seven sec squared three π‘₯ over five tan three π‘₯ with respect to π‘₯.

In this question, we are asked to integrate a quotient of trigonometric functions. Recall that the derivative of tan π‘₯ with respect to π‘₯ is sec squared π‘₯. Using the chain rule, we can deduce that for all constants π‘Ž, the derivative of tan π‘Žπ‘₯ with respect to π‘₯ is π‘Ž sec squared π‘Žπ‘₯. Therefore, the derivative of tan three π‘₯ with respect to π‘₯ is three sec squared three π‘₯. Using this, we note that the derivative of the function tan three π‘₯ in the denominator of the integrand of the integral we are asked to evaluate almost appears in the numerator. The difference is that instead of three sec squared three π‘₯, we have negative seven over five sec squared three π‘₯ in the numerator.

In order to evaluate this integral, we will use the following property. Given a nonzero function 𝑓 for all constants π‘Ž, the indefinite integral of π‘Ž multiplied by the derivative of 𝑓 with respect to π‘₯ 𝑓 prime of π‘₯ over 𝑓 of π‘₯ with respect to π‘₯ is equal to π‘Ž multiplied by the natural logarithm of the absolute value of 𝑓 of π‘₯ plus 𝐢, where 𝐢 is the constant of integration. In order to use this result, let 𝑓 of π‘₯ equal tan of three π‘₯. Then, the derivative of 𝑓 with respect to π‘₯, 𝑓 prime of π‘₯, is equal to three sec squared three π‘₯, as we have seen earlier.

We’re going to try to manipulate this equation in order to write the numerator of our integrand in the form π‘Ž multiplied by 𝑓 prime of π‘₯ for some constant π‘Ž. Dividing both sides of the equation by three, we obtain that one over three multiplied by 𝑓 prime of π‘₯ equals sec squared three π‘₯. Multiplying both sides of the resulting equation by negative seven over five, we obtain negative seven over 15 multiplied by 𝑓 prime of π‘₯ equals negative seven over five multiplied by sec squared of three π‘₯. Therefore, we can rewrite the integral we’re asked to determine as the indefinite integral of negative seven over 15 multiplied by 𝑓 prime of π‘₯ over 𝑓 of π‘₯ with respect to π‘₯, where 𝑓 of π‘₯ equals tan of three π‘₯.

Letting π‘Ž equal negative seven over 15 in the result described earlier, we obtain that the integral in question evaluates to negative seven over 15 multiplied by the natural logarithm of the absolute value of 𝑓 of π‘₯ plus 𝐢, where 𝐢 is the constant of integration. Substituting 𝑓 of π‘₯ equals tan of three π‘₯, we obtain that the integral in question is equal to negative seven over 15 multiplied by the natural logarithm of the absolute value of the function tan of three π‘₯ plus 𝐢, where 𝐢 is the constant of integration. This is our final answer.

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