Question Video: Integration Involving the Reciprocal Trigonometric Functions | Nagwa Question Video: Integration Involving the Reciprocal Trigonometric Functions | Nagwa

# Question Video: Integration Involving the Reciprocal Trigonometric Functions Mathematics • Third Year of Secondary School

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Determine β« sec ((2π₯ + 1)/3) tan ((2π₯ + 1)/3) dπ₯.

02:25

### Video Transcript

Determine the integral of the sec of two π₯ plus one over three multiplied by the tan of two π₯ plus one over three with respect to π₯.

In this question, weβre asked to evaluate the integral of a product of two trigonometric functions, the secant of an argument and the tangent of an argument. And this can immediately remind us of one of our standard integral results. The integral of the sec of π multiplied by the tan of π with respect to π is the sec of π plus the constant of integration πΆ. We canβt apply this result directly to this integral because we can see the argument of the secant function and the tangent function is more complicated. So instead, weβll integrate this by using a substitution. Weβll set π’ equal to the argument two π₯ plus one over three.

To apply integration by substitution, we recall we need to find an expression for the differentials. So weβre going to need to differentiate our substitution π’ with respect to π₯. And since π’ is a linear function, the derivative of π’ with respect to π₯ will be the coefficient of π₯. In this case, dπ’ by dπ₯ is equal to two-thirds. Now dπ’ by dπ₯ is not a fraction. However, we can treat it a little bit like a fraction to help us find an expression for the differentials. We get that dπ₯ is equal to three over two dπ’.

Weβre now ready to apply our substitution to help us evaluate our integral. Weβll replace two π₯ plus one over three with π’ and dπ₯ with three over two dπ’. This gives us the integral of the sec of π’ times the tan of π’ times three over two with respect to π’. And now this is almost in the form of our integral result. We just need to take the factor of three over two outside of our integral. We get three over two times the integral of sec π’ times tan π’ with respect to π’.

We then apply our integral rule. We get three over two times the sec of π’ plus the constant of integration πΆ, where itβs important to remember that πΆ is just a constant. So itβs not affected by the factor of three over two. Finally, since our original integral was in terms of π₯, we should rewrite this expression in terms of π₯. We need to use our π’-substitution, which gives us three over two times the sec of two π₯ plus one over three plus π, which is our final answer.

Therefore, by using a π’-substitution and one of our standard integral results, we were able to show the integral of the sec of two π₯ plus one over three times the tan of two π₯ plus one over three with respect to π₯ is equal to three over two times the sec of two π₯ plus one over three plus the constant of integration πΆ.

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