Trigonometric Integration: getting wrong function as answer

[morseterry60]

Hello, I'm working on trigonometric integration and for the second time I've gotten the answer right numerically, but with the wrong function. For example, in this equation I get sec in the answer whereas the book and WolframAlpha both get tan. Also the sign of my answer is different. Here is my work.

My answer is incorrect because tangent should be in place of secant, and the final answer is an addition, not a subtraction. What have I done wrong?

 

[Deleted member 4993]

Hello, I'm working on trigonometric integration and for the second time I've gotten the answer right numerically, but with the wrong function. For example, in this equation I get sec in the answer whereas the book and WolframAlpha both get tan. Also the sign of my answer is different. Here is my work.

My answer is incorrect because tangent should be in place of secant, and the final answer is an addition, not a subtraction. What have I done wrong?

Did you differentiate your answer to check whether you get back the original function?

 

[HallsofIvy]

You know, I presume, that \(\displaystyle sec^2(x)= tan^2(x)+ 1\). Replace each \(\displaystyle sec^2(x)\) with \(\displaystyle tan^2(x)+ 1\).

 

[Deleted member 4993]

I suspect the author of the textbook expected you to solve the problem in the following way:

\(\displaystyle \displaystyle{\int\dfrac{tan^3(x)}{cos^4(x)}dx}\)

\(\displaystyle = \displaystyle{\int tan^3(x)sec^4(x) dx}\)

\(\displaystyle = \displaystyle{\int tan^3(x)sec^2(x) [1 + tan^2(x)] dx}\)

\(\displaystyle = \frac{1}{4}tan^4(x) + \frac{1}{6}tan^6(x) + C \)