Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1-\cot x}{1+\cot x} d x$

Solution:

Convert $\cot x$ in form of $\sin x$ and $\cos x$.

$\Rightarrow \cot x=\frac{\cos x}{\sin x}$

$\therefore$ The equation now becomes

$\Rightarrow \int \frac{1-\frac{\cos x}{\sin x}}{1+\frac{\cos x}{\sin x}} d x$

$\Rightarrow \int \frac{\frac{\cos x-\sin x}{\sin x}}{\frac{\cos x+\sin x}{\sin x}} d x$

$\Rightarrow \int \frac{\cos x-\sin x}{\cos x+\sin x} d x$

Assume $\cos x+\sin x=t$

$\therefore \mathrm{d}(\cos x+\sin x)=\mathrm{dt}$

$=\cos x-\sin x$

$\therefore \mathrm{dt}=\cos x-\sin x$

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t}}$

$=\ln |t|+c$

But $t=\cos x+\sin x$

$\therefore \ln |\cos x+\sin x|+c$

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