Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:
\(\int \frac{dx}{1+2\sin x+\cos x}=\)
Options:
\(\log \left[1+2\tan \left(\frac{x}{2}\right)\right]+c\)
\(\log \left[1-2\tan \left(\frac{x}{2}\right)\right]+c\)
\(\frac{1}{2}\log \left[1+2\tan \left(\frac{x}{2}\right)\right]+c\)
None of the above
Correct Answer:
\(\frac{1}{2}\log \left[1+2\tan \left(\frac{x}{2}\right)\right]+c\)
Explanation:
\(\begin{aligned}\int \frac{dx}{11+2\sin x+\cos x}&=\int \frac{dx}{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}+2\times 2\sin \frac{x}{2}\cos \frac{x}{2}+\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}\\ &=\int \frac{\sec^{2}\frac{x}{2}}{2\left(1+2\tan \frac{x}{2}\right)}dx\\ &=\frac{1}{2}\int \frac{1}{t}dt \text{ where }t=1+2\tan \frac{x}{2}\\ &=\frac{1}{2}\log t+c\\ &=\frac{1}{2}\log \left(1+2\tan \frac{x}{2}\right)+c\end{aligned}\)