How do you evaluate the integral $\int \frac{1 - sin x}{cos x} d x$ $∫ (1-sinx)/cosx dx$ ?

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Answer 1 · 2015-02-20T20:53:06

Well, this one is tough and I'll need to use a trick.

I multiply and divide by $1 + sin (x)$ $1+sin(x)$ !!!
So you get:

$\int \frac{(1 - sin (x)) (1 + sin (x))}{cos (x) (1 + sin (x))} d x =$ $int[(1-sin(x))(1+sin(x))]/[cos(x)(1+sin(x))]dx=$

$= \int \frac{1 - {sin}^{2} (x)}{cos (x) (1 + sin (x))} d x =$ $=int[1-sin^2(x)]/[cos(x)(1+sin(x))]dx=$

$= \int \frac{{cos}^{2} (x)}{cos (x) (1 + sin (x))} d x =$ $=int[cos^2(x)]/[cos(x)(1+sin(x))]dx=$

$= \int \frac{cos (x)}{1 + sin (x)} d x =$ $=int[cos(x)]/[1+sin(x)]dx=$

but $d [sin (x)] = cos (x) d x$ $d[sin(x)]=cos(x)dx$

and:
$= \int \frac{1}{1 + sin (x)} d [sin (x)] =$ $=int[1]/[1+sin(x)]d[sin(x)]=$ and finally:

$= ln | 1 + sin (x) | + c$ $=ln|1+sin(x)|+c$

Hope it helps

How do you evaluate the integral ∫1−sinxcosxdx∫ (1-sinx)/cosx dx?