How do you find the antiderivative of $\int \frac{sin x}{{cos}^{3} x} d x$ $int sinx/cos^3x dx$ ?

Question

17887 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-26T11:11:43

The answer is $= \frac{1}{2 {cos}^{2} x} + C$ $=1/(2cos^2x)+C$

We need

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (x \neq 0)$ $intx^ndx=x^(n+1)/(n+1)+C(x!=0)$

$(cosx)'=-sinx$

We perform this integral by substitution

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

Therefore,

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

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

How do you find the antiderivative of int sinx/cos^3x dx∫sinxcos3xdxint sinx/cos^3x dx?