Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Find the integral: $\int \frac{1}{1+\tan x} d x$
✓
Answer
$\int \frac{d x}{1+\tan x}=\int \frac{\cos x d x}{\cos x+\sin x}$ = $\frac{1}{2} \int \frac{(\cos x+\sin x+\cos x-\sin x) d x}{\cos x+\sin x}$ = $\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x$ = $\frac{x}{2}+\frac{C_{1}}{2}+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x$ ...(i) Now, consider I = $\int \frac{\cos x-\sin x}{\cos x+\sin x} d x$ Put cos x + sin x = t so that (cos x - sin x) dx = dt Therefore, $\mathrm{I}=\int \frac{d t}{t}=\log |t|+\mathrm{C}_{2}=\log |\cos x+\sin x|+\mathrm{C}_{2}$ Putting it in (i), we get $\int \frac{d x}{1+\tan x}=\frac{x}{2}+\frac{\mathrm{C}_{1}}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{\mathrm{C}_{2}}{2}$ = $\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{C_{1}}{2}+\frac{C_{2}}{2}$ = $\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\mathrm{C},\left(\mathrm{C}=\frac{\mathrm{C}_{1}}{2}+\frac{\mathrm{C}_{2}}{2}\right)$
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