Integral Calculus
86438
\(\int_{0}^{\pi / 2} \frac{d x}{5+4 \cos x}=\)
1 \(\tan ^{-1}\left(\frac{1}{3}\right)\)
2 \(\frac{2}{3} \tan ^{-1}\left(\frac{1}{3}\right)\)
3 \(2 \tan ^{-1}\left(\frac{1}{3}\right)\)
4 \(\frac{1}{3} \tan ^{-1}\left(\frac{1}{3}\right)\)
Explanation:
(B) : Given,
\(I =\int_{0}^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x}\)
\(=\int_{0}^{\frac{\pi}{2}} \frac{d x}{5+4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)}\left(\because \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)\)
\(=\int_{0}^{\frac{\pi}{2}} \frac{\left(1+\tan ^{2} \frac{x}{2}\right) d x}{5\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)}\)
\(=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2} d x}{9+\tan ^{2} \frac{x}{2}}\)
Let, \(\quad \tan \frac{\mathrm{x}}{2}=\mathrm{t}\)
\(\frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t \Rightarrow \sec ^{2} \frac{x}{2} d x=2 d t\)
When, \(\quad \mathrm{x}=0, \quad \mathrm{t}=0\)
\(x=\frac{\pi}{2}, t=\tan \frac{\pi}{4}=1\)
Now,
\(\mathrm{I} =\int_{0}^{\frac{\pi}{2}} \frac{1}{\mathrm{t}^{2}+9} \cdot 2 \mathrm{dt}=2 \int_{0}^{1} \frac{1}{\mathrm{t}^{2}+3^{2}} \mathrm{dt}\)
\(=2\left[\frac{1}{3} \tan ^{-1} \frac{\mathrm{t}}{3}\right]_{0}^{1}=\frac{2}{3}\left[\tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1} 0\right]_{0}^{1}\)
\(=\frac{2}{3} \tan ^{-1}\left(\frac{1}{3}\right)\)