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

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Answer 1 · 2015-02-18T14:17:41

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 $-intx^-4dx$ )

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

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