$\int \frac{d x}{\tan x+\cot x+\sec x+\operatorname{cosec} x}=$
$\int \frac{d x}{\tan x+\cot x+\sec x+\operatorname{cosec} x}=$
- $\frac{1}{2}(\sin x-\cos x+x)+c$
- $\frac{1}{2}(\sin x-\cos x-\tan x+\cot x)+c$
- $\frac{1}{2}(\sin x-\cos x-x)+c$
- $\frac{1}{2}(\sin x+\cos x-\tan x-\cot x)+c$
Solution
We have,
$\begin{aligned}
& \int \frac{d x}{\tan x+\cot x+\sec x+\operatorname{cosec} x} \\
= & \int \frac{d x}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}+\frac{1}{\cos x}+\frac{1}{\sin x}} \\
= & \int \frac{\sin x \cos x d x}{\sin ^2 x+\cos ^2 x+\sin x+\cos x} \\
= & \int \frac{\sin x \cos x d x}{1+\sin x+\cos x}
\end{aligned}$
multiply and divide by 2
$\begin{aligned}
& =\frac{1}{2} \int \frac{2 \sin x \cos x}{1+\sin x+\cos x} d x \\
& =\frac{1}{2} \int \frac{2 \sin x \cos x+1-1}{1+\sin x+\cos x} d x \\
& =\frac{1}{2} \int \frac{\left(\sin ^2 x+\cos ^2 x+2 \sin x \cos x\right)-1}{1+\sin x+\cos x} d x \\
& =\frac{1}{2} \int \frac{(\sin x+\cos x)^2-1}{1+\sin x+\cos x} d x \\
& =\frac{1}{2} \int \frac{(\sin x+\cos x+1)(\sin x+\cos x-1)}{1+\sin x+\cos x} d x \\
& =\frac{1}{2} \int(\sin x+\cos x-1) d x \\
& =\frac{1}{2}[-\cos x+\sin x-x]+c \\
& =\frac{1}{2}[\sin x-\cos x-x]+c
\end{aligned}$
Asked in: MHT CET Full Test 7
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