If int frac{1}{1-cot x} dx = Ax + B log |sin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int \frac{1}{1-\cot x} dx$. Substitute $\cot x = \frac{\cos x}{\sin x}$,so $I = \int \frac{1}{1-\frac{\cos x}{\sin x}} dx = \int \frac{\

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $I = \int \frac{1}{1-\cot x} dx$. Substitute $\cot x = \frac{\cos x}{\sin x}$,so $I = \int \frac{1}{1-\frac{\cos x}{\sin x}} dx = \int \frac{\sin x}{\sin x - \cos x} dx$. Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2 \sin x}{\sin x - \cos x} dx$. Rewrite the numerator: $2 \sin x = (\sin x - \cos x) + (\sin x + \cos x)$. Thus,$I = \frac{1}{2} \int \frac{(\sin x - \cos x) + (\sin x + \cos x)}{\sin x - \cos x} dx = \frac{1}{2} \int \left( 1 + \frac{\sin x + \cos x}{\sin x - \cos x} \right) dx$. Let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) dx$. $I = \frac{1}{2} \left( \int 1 dx + \int \frac{1}{u} du \right) = \frac{1}{2} (x + \log |u|) + C = \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C$. Comparing with $Ax + B \log |\sin x - \cos x| + C$,we get $A = \frac{1}{2}$ and $B = \frac{1}{2}$. Therefore,$A + B = \frac{1}{2} + \frac{1}{2} = 1$.

If $\int \frac{1}{1-\cot x} dx = Ax + B \log |\sin x - \cos x| + C$,then $A + B = \dots$

Similar Questions

Integrate the function: $\frac{5 x-2}{1+2 x+3 x^{2}}$

If $I_n = \int \frac{1}{(x^2+1)^n} dx$,then $2n I_{n+1} - (2n-1) I_n = $

If $\int \frac{\sin \theta}{\sin 3 \theta} d \theta = \frac{1}{2 k} \log \left|\frac{k+\tan \theta}{k-\tan \theta}\right|+c$,then $k=$

Let $I_n = \int \sec^n x \, dx$. If $5 I_6 - 4 I_4 = f(x)$,then $f\left(\frac{\pi}{4}\right)$ is equal to

$\int \frac{1}{(x^2+1)^2} dx = . . . . . .$

Vedclass Test Series

Exam Paper Generator

Online Exam Module