Solveeit Logo
Login

Question

Question: \(\log(x\sin x - \cos x) + c\)...

log⁡(xsin⁡x−cos⁡x)+c\log(x\sin x - \cos x) + c

A

log⁡(sin⁡x−xcos⁡x)+c\log(\sin x - x\cos x) + c

B

∫sin⁡2x1+sin⁡2xdx=\int_{}^{}{\frac{\sin 2x}{1 + \sin^{2}x}dx =}

C

logsin2x+c{logsin}2x + c

D

log⁡(1+sin⁡2x)+c\log(1 + \sin^{2}x) + c

Answer

∫sin⁡2x1+sin⁡2xdx=\int_{}^{}{\frac{\sin 2x}{1 + \sin^{2}x}dx =}

Explanation

Solution

∫e−xcosec2(2e−x+5) 6mudx=\int_{}^{}{e^{- x}\text{cose}\text{c}^{2}(2e^{- x} + 5)}\mspace{6mu} dx =

12cot⁡(2e−x+5)+c\frac{1}{2}\cot(2e^{- x} + 5) + c

Put −12cot⁡(2e−x+5)+c- \frac{1}{2}\cot(2e^{- x} + 5) + c

then it reduced to 2cot⁡(2e−x+5)+c2\cot(2e^{- x} + 5) + c.

Trick : By inspection,

−2cot⁡(2e−x+5)+c- 2\cot(2e^{- x} + 5) + c

∫sin⁡3x 6mu. 6mucos⁡x 6mudx=\int_{}^{}{\sin^{3}x\mspace{6mu}.\mspace{6mu}\cos x\mspace{6mu} dx =}.