Question

integration of sinxcosx

Answer 1

[tex] \int\ sinx cosx dx [/tex] 

Multiplying and dividing sinxcosx with 2

We get 2sinxcosx/2

But we know tht 2sinxcosx=sin2x

$\int\sin(nx)dx=-cosnx/n$
 
So,
$\int\sin2xdx/2$
$\int\sin2xdx/2=-cos2x/4$

HOPE THIS HELPED U.......

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\large\boxed{\sf{\int \:sinxcosx\:dx}}$

♣ ᴀɴꜱᴡᴇʀ :

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

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\text { Apply u - substitution: } u=\sin (x)$

$=\int \:udu$

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

$=\dfrac{u^{1+1}}{1+1}$

$\mathrm{Substitute\:back}\:u=\sin \left(x\right)$

$=\dfrac{\sin ^{1+1}\left(x\right)}{1+1}$

$\text { Simplify } \dfrac{\sin ^{1+1}(x)}{1+1}: \dfrac{1}{2} \sin ^{2}(x)$

$=\dfrac{1}{2}\sin ^2\left(x\right)$

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

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