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How do I evaluate the integral   #intsec^3(x) tan(x) dx#?

Answer 1

Isaiah Allen

I would start by writing your integrand as:

#int1/cos^3(x)*sin(x)/cos(x)dx=intsin(x)/cos^4(x)dx=#

Now: #d[cos(x)]=-sin(x)dx#

I can write the integral in a new equivalent form:

#-int(d[cos(x)])/cos^4(x)=# now you can use #cos(x)# as if it were a simple #x# during your integration, giving:

#-int(d[cos(x)])/cos^4(x)=-intcos^(-4)(x)d[cos(x)]=# (as for #-intx^-4dx#)

#=cos^(-3)/3+c=1/(3cos^3(x))+c#

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Answer 2

Samantha Adkison

To evaluate the integral (\int \sec^3(x) \tan(x) , dx), you can use the substitution method. Let ( u = \sec(x) + \tan(x) ) and ( du = (\sec(x)\tan(x) + \sec^2(x)) , dx ).

Then, the integral becomes (\int u^3 , du), which is straightforward to integrate. After integrating, substitute back ( u = \sec(x) + \tan(x) ) to get the final result.

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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.

How do I evaluate the integral&nbsp;&nbsp; #intsec^3(x) tan(x) dx#?

How do I evaluate the integral #intsec^3(x) tan(x) dx#?