Integral Calculus
86475
\(\int \frac{\operatorname{cosec} x}{\cos ^{2}\left(1+\log \tan \frac{x}{2}\right)} d x=\)
1 \(\sin ^{2}\left[1+\log \tan \frac{x}{2}\right]+C\)
2 \(\tan \left[1+\log \tan \frac{x}{2}\right]+C\)
3 \(\sec ^{2}\left[1+\log \tan \frac{x}{2}\right]+C\)
4 \(-\tan \left[1+\log \tan \frac{\mathrm{x}}{2}\right]+\mathrm{C}\)
Explanation:
(B) : \(I=\int \frac{\operatorname{cosec} x}{\cos ^{2}\left(1+\log \tan \frac{x}{2}\right)} d x\)
Let, \(\quad 1+\log \tan \frac{\mathrm{x}}{2}=\mathrm{t}\)
\(\frac{1}{\tan \frac{x}{2}} \cdot \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \Rightarrow \frac{\cos \frac{x}{2}}{2 \sin \frac{x}{2}} \times \frac{1}{\cos ^{2} \frac{x}{2}} d x=d t\)
\(\frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} d x=d t \Rightarrow \frac{1}{\sin x} d x=d t\)
\(\operatorname{cosec} x \cdot d x=d t\)
So, \(\quad I=\int \frac{1}{\cos ^{2} t} d t\)
\(I=\int \sec ^{2} t d t\)
\(\mathrm{I}=\tan \mathrm{t}+\mathrm{C}\)
\(I=\tan \left[1+\log \tan \frac{x}{2}\right]+C\)