Question

2+3cosx/sin²x,Integrate the given function w.r.t. x considering them well defined and integrable over proper domain.

Answer 1

we have to find the value of $\int{\frac{2+3cosx}{sin^2x}}\,dx$

$\int{\frac{2+3cosx}{sin^2x}}\,dx$

$=\int{\frac{2}{sin^2x}}\,dx+\int{\frac{3cosx}{sin^2x}},dx$

$=2\int{cosec^2x}\,dx+3\int{cotx.cosecx}\,dx$

$=2[-cotx]+3[-cosecx]+C$

$=-(2cotx+3cosecx)+C$

$=-\frac{2cosx+3}{sinx}+C$

Answer 2

Dear Student,

Answer:
$\int \frac{2 + 3 \: cos \: x}{ {sin}^{2}x } dx = - 2 \: cot \: x + 3 \: cosec \: x + c$

Solution:

$\int\frac{2 + 3 \: cos \: x}{ {sin}^{2}x } dx \\ \\ \int\frac{2}{ {sin}^{2}x } dx + \frac{3 \: cos \: x}{ {sin}^{2} x} dx \\ \\ 2\int {cosec}^{2} x \: dx + 3\int\frac{cos \: x}{ {sin}^{2}x } dx$
Since
$\int{cosec}^{2} x \: dx = - cot \: x + c$
$2\int {cosec}^{2} x \: dx = - 2 \: cot \: x + c$

$3\int \frac{cos \: x}{ {sin}^{2}x } \: dx =$
let sin x = t

- cos x dx = dt

so substitute

$- 3 \int\frac{1}{ {t}^{2} } dt = - 3 \frac{ {t}^{ - 2 + 1} }{ - 2 + 1} + c \\ \\ = \frac{3}{t} + c \\ \\ undo \: substitution \\ \\ = \frac{3}{sin \: x} + c \\ \\ = 3 \: cosec \: x + c$
add both integral

$= - 2 \: cot \: x + 3 \: cosec \: x + c$
Hope it helps you.