Solveeit Logo
Login

Question

Question: \[\int_{}^{}{\frac{dx}{1 + \cot x} =}\]...

∫dx1+cot⁡x=\int_{}^{}{\frac{dx}{1 + \cot x} =}

A

12x+12log⁡∣sin⁡x+cos⁡x∣+c\frac{1}{2}x + \frac{1}{2}\log|\sin x + \cos x| + c

B

12x−12log⁡∣sin⁡x+cos⁡x∣+c\frac{1}{2}x - \frac{1}{2}\log|\sin x + \cos x| + c

C

−12x+12log⁡∣sin⁡x+cos⁡x∣+c\frac{- 1}{2}x + \frac{1}{2}\log|\sin x + \cos x| + c

D

None of these

Answer

12x−12log⁡∣sin⁡x+cos⁡x∣+c\frac{1}{2}x - \frac{1}{2}\log|\sin x + \cos x| + c

Explanation

Solution

Here,I=∫dx1+cot⁡x=∫dx1+cos⁡xsin⁡x=∫sin⁡xsin⁡x+cos⁡xdxI = \int_{}^{}\frac{dx}{1 + \cot x} = \int_{}^{}{\frac{dx}{1 + \frac{\cos x}{\sin x}} = \int_{}^{}\frac{\sin x}{\sin x + \cos x}dx}Let sin⁡x=Mddx(sin⁡x+cos⁡x)+N(sin⁡x+cos⁡x)\sin x = M\frac{d}{dx}(\sin x + \cos x) + N(\sin x + \cos x)

⇒\Rightarrow sin⁡x=M(cos⁡x−sin⁡x)+N(sin⁡x+cos⁡x)\sin x = M(\cos x - \sin x) + N(\sin x + \cos x) Comparing the coefficients of sin⁡x\sin x and cos⁡x\cos x of both the sides, we have 1=−M+Nand0=M+N1 = - M + N\text{and}0 = M + NSolving these equations, we have M=−12andN=12M = \frac{- 1}{2}\text{and}N = \frac{1}{2}

∴\therefore sin⁡x=−12(cos⁡x−sin⁡x)+12(sin⁡x+cos⁡x)\sin x = - \frac{1}{2}(\cos x - \sin x) + \frac{1}{2}(\sin x + \cos x) Hence,I=∫−12(cos⁡x−sin⁡x)+12(sin⁡x+cos⁡x)sin⁡x+cos⁡xdx=−12∫cos⁡x−sin⁡xsin⁡x+cos⁡xdx+12∫1dx=−12log⁡∣sin⁡x+cos⁡x∣+12x+c.I = \int_{}^{}\frac{\frac{- 1}{2}(\cos x - \sin x) + \frac{1}{2}(\sin x + \cos x)}{\sin x + \cos x}dx = - \frac{1}{2}\int_{}^{}\frac{\cos x - \sin x}{\sin x + \cos x}dx + \frac{1}{2}\int_{}^{}{1dx} = - \frac{1}{2}\log|\sin x + \cos x| + \frac{1}{2}x + c.