Question

integrate by substitution easily

Answer 1

HELLO DEAR,

GIVEN:-
∫sinx/(sinx - cosx) . dx

⇒1/2* ∫2sinx/(sinx - cosx).dx

⇒1/2 * ∫$\bold{\frac{(sinx + cosx) + (sinx - cosx)}{(sinx - cosx)}\,dx}$

⇒1/2 * $[\bold{\int{\frac{cosx + sinx}{sinx - cosx}}\,dx + \int{\frac{sinx - cosx}{sinx - cosx}}\,dx}]$

let t = sinx - cosx
⇒dt = dx.(cosx + sinx)

So , ⇒1/2[dt/t + 1dx]

⇒1/2 * (log|t| + x) + c

⇒ 1/2 * log|sinx - cosx| + 1/2x + c

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

SOLUTION:-

$\implies \int \: \frac{ \sin(x) }{( \sin(x) - \cos(x) } dx$

multiplie and divide 1/2

$\implies \frac{1}{2} \int \: \frac{2 \sin(x) }{( \sin(x) - \cos(x) ) }$

can we be written as

$\implies \: \frac{1}{2 } \int \: \frac{ \sin \: x+ \sin\: x }{( \sin x - \cos \: x ) }$

add and subscrate cos x

$\implies \: \frac{1}{2} \int \: \frac{ (\sin \: x + \cos \: x) + ( \sin \: x - \cos \: x)}{( \sin \: x - \cos \: x) }$

can be written as

$\implies \: \frac{1}{2} \int \: \frac{ \sin \: x + \cos \: x }{ \sin \: x - \cos \: x } + \frac{1}{2} \int \cancel{ \frac{ \sin \: x - \cos \: x }{ \sin x - \cos \: x } }$

$let \frac{1}{2} \int \: \frac{ \sin(x) + \cos(x) }{ \sin(x) - \cos(x) } = i_{}$

and

$\implies \frac{1}{2} \int \: 1 \: dx = ii$

solve I

let sin x -cos x=t----(1)

$\implies \: \frac{d}{dx} \sin(x) - \frac{d}{dx} \cos(x) = dt \\ \\ \implies \: \cos(x) - ( - \sin(x) ) = dt \\ \\ \implies \: \cos(x) + \sin(x) = dt$

put value I

$\implies \: \frac{1}{2} \int \: \frac{1}{t} dt \\ \\ \implies \: \frac{1}{2} log \: t$

put t value

$\implies \: \frac{1}{2} log \: | \cos(x) - \sin(x) |$

solve ii

$\implies \: \frac{1}{2} x$

add I +ii

$\implies \: \frac{1}{2} \: log( | \cos(x) - \sin(x) | ) + \frac{1}{2} x$+c

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I hope it helps you