Question

If y=log(x^x + cosec²x)
Prove that:
$\frac{dy}{dx} = [{x}^{x} (1+logx) - 2cosec²x.cotx]$
​

Answer 1

Let y = (log x)x – (cos x)cotx. Put u = (log x)x and v = (cos x)cotx. Then y = u – v ∴ `"dy"/"dx" = "du"/"dx" - "dv"/"dx"` ...(1) Take u = (log x)x

Answer 2

I=integ. of cosec x.dx

Multiplying by (cosec x-cot x) in Nr and Dr.

I=integ. of cosec x.(cosec x- cot x).dx/(cosec x-cot x).

Let cosec x- cot x = t

(-cosec x.cot x +cosec^2x ).dx = dt

cosec x.(-cot x+ cosec x).dx= dt

I=integ. of 1/t.dt.

I= log | t |+ C.

I= log |cosec x-cot x| +C……………….(1)

For second form:-

I = log | 1/sin x - cos x /sin x | + C

I= log |(1-cos x)/sin x | +C.

I= log |(1–1+2sin^2 x/2)/(2.sin x/2.cos x/2)| +C.

I= log |(2sin^2 x/2)/(2.sin x/2.cos x/2) | +C.

I= log |sin x/2/cos x/2| + C.

I= log | tan x/2 | + C…………………….(2). Answer.

please mark me as brainlist And follow

......