MHT CET · Maths · Definite Integration
\(\int_0^\pi \frac{\mathrm{d} x}{4+3 \cos x}=\)
- A \(\frac{2 \pi}{7}\)
- B \(\frac{\pi}{\sqrt{7}}\)
- C \(\frac{\pi}{2 \sqrt{7}}\)
- D \(\frac{\pi}{7}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{\sqrt{7}}\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{I}=\int_0^\pi \frac{\mathrm{dx}}{4+3 \cos x}\)
Put \(\tan \frac{x}{2}=\mathrm{t}\)
\(\therefore \mathrm{d} x=\frac{2 \mathrm{dt}}{1+\mathrm{t}^2}\) and \(\cos x=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\)
\(\therefore \mathrm{I}=\int_0^\pi \frac{\mathrm{d} x}{4+3 \cos x} =\int_0^{\infty} \frac{2 \mathrm{dt}}{7+\mathrm{t}^2} \)
\( =\left[\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\mathrm{t}}{\sqrt{7}}\right)\right]_0^{\infty} \)
\( =\frac{2}{\sqrt{7}}\left[\tan ^{-1} \infty-0\right] \)
\( =\frac{2}{\sqrt{7}} \cdot \frac{\pi}{2}=\frac{\pi}{\sqrt{7}}\)
Put \(\tan \frac{x}{2}=\mathrm{t}\)
\(\therefore \mathrm{d} x=\frac{2 \mathrm{dt}}{1+\mathrm{t}^2}\) and \(\cos x=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\)
\(\therefore \mathrm{I}=\int_0^\pi \frac{\mathrm{d} x}{4+3 \cos x} =\int_0^{\infty} \frac{2 \mathrm{dt}}{7+\mathrm{t}^2} \)
\( =\left[\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\mathrm{t}}{\sqrt{7}}\right)\right]_0^{\infty} \)
\( =\frac{2}{\sqrt{7}}\left[\tan ^{-1} \infty-0\right] \)
\( =\frac{2}{\sqrt{7}} \cdot \frac{\pi}{2}=\frac{\pi}{\sqrt{7}}\)
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