MHT CET · Maths · Definite Integration
If \(I_n=\int_0^{\frac{\pi}{4}} \tan ^n \theta d \theta\), then \(I_{12}+I_{10}=\)
- A \(\frac {1}{8}\)
- B \(\frac {1}{12}\)
- C \(\frac {1}{11}\)
- D \(\frac {1}{10}\)
Answer & Solution
Correct Answer
(C) \(\frac {1}{11}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int_0^{\frac{\pi}{4}}\left(\tan ^{\mathrm{n}} x+\tan ^{\mathrm{n}-2} x\right) \mathrm{d} x=\frac{1}{\mathrm{n}-1} \\ & \therefore \quad \mathrm{I}_{12}+\mathrm{I}_{10}=\int_0^{\frac{\pi}{4}}\left(\tan ^{12} \theta+\tan ^{10} \theta\right) \mathrm{d} \theta \\ &=\frac{1}{12-1} \\ &=\frac{1}{11}\end{aligned}\)
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