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How do you find the antiderivative of #int sinx/cos^3x dx#?

Answer 1

Emily Ammerman

The answer is #=1/(2cos^2x)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C(x!=0)#

#(cosx)'=-sinx#

We perform this integral by substitution

Let #u=cosx#, #=>#, #du=-sinxdx#

Therefore,

#int(sinxdx)/(cos^3x)=int(-du)/u^3#

#=1/(2u^2)=1/(2cos^2x)+C#

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Answer 2

Emilia Aguilar

To find the antiderivative of ∫(sinx/cos^3x) dx, you can use the substitution method. Let u = cos(x), then du = -sin(x) dx. Substituting, the integral becomes ∫-1/(u^3) du. Integrating, you get (-1/(-2u^2)) + C = 1/(2u^2) + C. Substituting back for u, you get 1/(2cos^2(x)) + C. So, the antiderivative of the given expression is 1/(2cos^2(x)) + C.

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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.