What is the integral of $\int \frac{sin x}{{cos}^{2} x} d x$ $int (sinx)/(cos^2x) dx$ ?

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Answer 1 · 2016-03-13T01:24:41

It is

$int (sinx)/(cos^2x) dx= int [1/cosx]'dx=1/cosx+c=secx+c$

Answer 2 · 2016-03-17T03:18:32

$secx+C$

We should try to use substitution by setting $u=cosx$ , so $du=-sinxdx$ .

This gives us the integral:

$intsinx/cos^2xdx=-int(-sinx)/cos^2xdx=-int1/u^2du=-intu^-2du$

From here, use the rule

$intu^ndu=u^(n+1)/(n+1)+C$

Thus,

$-intu^-2du=-u^(-1)/(-1)+C=1/u+C$

$=1/cosx+C=secx+C$

Answer 3 · 2016-03-17T22:06:26

$secx+C$

Alternatively, you could rewrite this in terms of other trigonometric functions:

$intsinx/cos^2xdx=int(1/cosx)(sinx/cosx)dx=intsecxtanxdx$

If you're familiar with the fact that $d/dx(secx)=secxtanx$ , then you'll recognize this common integral formula:

$intsecxtanxdx=secx+C$

What is the integral of int (sinx)/(cos^2x) dx∫sinxcos2xdxint (sinx)/(cos^2x) dx?