KCET · Maths · Definite Integration
If \(I_{n}=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x\), where \(n\) is positive integer, then \(I_{10}+I_{8}\) is equal to
- A 9
- B \(\frac{1}{7}\)
- C \(\frac{1}{8}\)
- D \(\frac{1}{9}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{9}\)
Step-by-step Solution
Detailed explanation
\(I_{n}=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x...(i)\)
\(I_{n+2}=\int_{0}^{\frac{\pi}{4}} \tan ^{n+2} x d x...(ii)\)
Adding Eqs. (i) and (ii),
\(\begin{aligned}
I_{n}+I_{n+2} &=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x+\int_{0}^{\frac{\pi}{4}} \tan ^{n+2} x d x \\
&=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x \cdot \sec ^{2} x d x \\
&=\left[\frac{\tan ^{n+1} x}{n+1}\right]_{0}^{\frac{\pi}{4}}=\frac{1}{n+1}
\end{aligned}\)
Thus, \(I_{n}+I_{n+2}=\frac{1}{(n+1)}\)
Substitute 8 for \(x\),
\(I_{8}+I_{10}=\frac{1}{9}\)
\(I_{n+2}=\int_{0}^{\frac{\pi}{4}} \tan ^{n+2} x d x...(ii)\)
Adding Eqs. (i) and (ii),
\(\begin{aligned}
I_{n}+I_{n+2} &=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x+\int_{0}^{\frac{\pi}{4}} \tan ^{n+2} x d x \\
&=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x \cdot \sec ^{2} x d x \\
&=\left[\frac{\tan ^{n+1} x}{n+1}\right]_{0}^{\frac{\pi}{4}}=\frac{1}{n+1}
\end{aligned}\)
Thus, \(I_{n}+I_{n+2}=\frac{1}{(n+1)}\)
Substitute 8 for \(x\),
\(I_{8}+I_{10}=\frac{1}{9}\)
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