1 1+ dx= - Vidyadip

MCQ

$\int\frac{1}{1+\tan\text{x}}\text{ dx}=$

A
$\log_\text{e}(\text{x}+\sin\text{x})+\text{C}$
B
$\log_\text{e}(\sin\text{x}+\cos\text{x})+\text{C}$
C
$2\sec^2\frac{\text{x}}{2}+\text{C}$
✓
$\frac{1}{2}\big[\text{x}+\log(\sin\text{x}+\cos\text{x})\big]+\text{C}$

✓

Answer

Correct option: D.

$\frac{1}{2}\big[\text{x}+\log(\sin\text{x}+\cos\text{x})\big]+\text{C}$

$\text{I}=\int\frac{1}{1+\tan\text{x}}\text{ dx}$
$=\int\frac{\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Numerator can be written as,
$\cos\text{x}=\text{A}(\sin\text{x}+\cos\text{x})+\text{B}\frac{\text{d}(\sin\text{x}+\cos\text{x})}{\text{dx}}$
$\cos\text{x}=(\text{A}-\text{B})\sin\text{x}+(\text{A}+\text{B})\cos\text{x}$
$\text{A}-\text{B}=0$ and $\text{A}+\text{B}=1$
$\text{A}=\frac{1}{2}=\text{B}$
$\text{I}=\int\frac{\big[\frac{1}{2}(\sin\text{x}+\cos\text{x})+\frac{1}{2}(\cos\text{x}-\sin\text{x})\big]\text{dx}}{\sin\text{x}+\cos\text{x}}$
$\text{I}=\frac{1}{2}\int\Big(1+\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\text{I}=\frac{1}{2}\big[1+\int(\sin\text{x}+\cos\text{x})\big]+\text{C}$

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