Integrate the functions in Exercises: 1 1-

Question

Integrate the functions in Exercises:
$\frac{1}{1-\tan\text{x}}$

✓

Answer

$\text{Let I}=\int\frac{1}{1-\tan\text{x}}\text{ dx} =\int\frac{1}{1-\frac{\sin\text{x}}{\cos\text{x}}}\text{ dx} $
$=\int\frac{1}{\bigg(\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}}\bigg)}\text{ dx} =\int\frac{\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{2\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} =\frac{1}{2}\int\frac{\cos\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} $
Adding and subtracting $\sin\text{x}$ in the numerator,
$=\frac{1}{2}\int\frac{\cos\text{x}-\sin\text{x}+\sin\text{x}+\cos\text{x}} {\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{(\cos\text{x}-\sin\text{x})+(\sin\text{x}+\cos\text{x})}{\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\bigg(\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}-\sin\text{x}}-\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\int\bigg(1-\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\bigg)\text{ dx} $
$\frac{1}{2}\bigg[\int1\text{ dx}-\int\frac{-\sin\text{x}-\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx}\bigg]$
$=\frac{1}{2}\big[\text{x}-\log\begin{vmatrix}\cos\text{x} - \sin \text{x} \end{vmatrix}\big]+\text{c} $

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "1 Marks" from INTEGRALS→INTEGRALS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

INTEGRALS — Maths STD 12 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions