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How do I evaluate #inttan (x) sec^3(x) dx#?

Answer 1

I can write the integral as: #intSin(x)/cos(x)*1/cos^3(x)dx=# #=intsin(x)/cos^4(x)dx=# but: #d(cos(x))=-sin(x)dx# and the integral becomes (multiplying and dividing by #-1#): #=int(-sin(x))/-cos^4(x)dx=# #=int1/-cos^4(x)d(cos(x)=# #=-intcos^(-4)(x)d(cos(x))=# #=-cos^(-3)(x)/(-3)+c=# #=cos^(-3)(x)/(3)+c#

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Answer 2

Ezra Adams

To evaluate the integral ∫tan(x) sec^3(x) dx, use substitution method. Let u = sec(x), then du = sec(x)tan(x) dx. The integral becomes ∫u^2 du. Integrate u^2 to get (1/3)u^3. Substitute back u = sec(x) to get (1/3)sec^3(x) + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.