Question

simplify this integration​

Answer 1

Answer:

-1/2.cotx+c

Step-by-step explanation:

Easy steps:-

1) Cos2x=sin²x-sin2x

2) sin²x+cos²x=1

Substituting (2) in eqn (1).

and putting it in (1-cos2x) While solving,

cancelling the square terms and substituting.

Let it be turning to cotx

taking out 1/2 common bur due to transmision of any kind its negative

So final comes is -1/2cotx+c

Refer attachment !!

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\boxed{\sf{\int \dfrac{1}{1-cos2x}dx}}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\int \dfrac{1}{1-\cos \left(2x\right)}dx=-\dfrac{1}{2\tan \left(x\right)}+C}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\text { Apply u - substitution: } u=2 x$

$=\int \dfrac{1}{2\left(1-\cos \left(u\right)\right)}du$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=\dfrac{1}{2}\cdot \int \dfrac{1}{1-\cos \left(u\right)}du$

$\text { Apply } u-\text { substitution: } v=\tan \left(\dfrac{u}{2}\right)$

$=\dfrac{1}{2}\cdot \int \dfrac{1}{v^2}dv$

$\mathrm{Apply\:exponent\:rule}:\quad \dfrac{1}{a^b}=a^{-b}$

$\dfrac{1}{v^2}=v^{-2}$

$=\dfrac{1}{2}\cdot \int \:v^{-2}dv$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\dfrac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$=\dfrac{1}{2}\cdot \dfrac{v^{-2+1}}{-2+1}$

$\text { Substitute back }$

$=\dfrac{1}{2}\cdot \dfrac{\tan ^{-2+1}\left(\dfrac{2x}{2}\right)}{-2+1}$

$\text { Simplify } \dfrac{1}{2} \cdot \dfrac{\tan ^{-2+1}\left(\dfrac{2 x}{2}\right)}{-2+1}:-\dfrac{1}{2 \tan (x)}$

$=-\dfrac{1}{2\tan \left(x\right)}$

$\bf{Add\:a\:constant\:to\:the\:solution}$

$\large\boxed{\sf{=-\dfrac{1}{2\tan \left(x\right)}+C}}$