Explanation:
(D) : Given,
\(\mathrm{I}_{\mathrm{n}}=\int_{0}^{\pi / 4} \tan ^{\mathrm{n}} \theta \mathrm{d} \theta\)
Then, we have to find \(\mathrm{I}_{8}+\mathrm{I}_{6}\).
So, we can write-
\(\mathrm{I}_{8}=\int_{0}^{\pi / 4} \tan ^{8} \theta \mathrm{d} \theta\)
And, \(I_{6}=\int_{0}^{\pi / 4} \tan ^{6} \theta d \theta\)
\(\therefore \quad \mathrm{I}_{8}+\mathrm{I}_{6}=\int_{0}^{\pi / 4} \tan ^{8} \theta \mathrm{d} \theta+\int_{0}^{\pi / 4} \tan ^{6} \theta \mathrm{d} \theta\)
\(I_{8}+I_{6}=\int_{0}^{\pi / 4}\left(\tan ^{8} \theta d \theta+\tan ^{6} \theta d \theta\right)\)
\(\mathrm{I}_{8}+\mathrm{I}_{6}=\int_{0}^{\pi / 4} \tan ^{6} \theta\left(\tan ^{2} \theta+1\right) \mathrm{d} \theta\)
\(I_{8}+I_{6}=\int_{0}^{\pi / 4} \tan ^{6} \theta \sec ^{2} \theta d \theta \quad\left(\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right)\)
Let, \(\quad \tan \theta=\mathrm{t}\)
\(\sec ^{2} \theta \mathrm{d} \theta=\mathrm{dt}\)
For limit-
When, \(\theta=0, t=0\)
When, \(\theta=\frac{\pi}{4}\), \(\mathrm{t}=1\)
\(\therefore \quad \mathrm{I}_{8}+\mathrm{I}_{6}=\int_{0}^{1} \mathrm{t}^{6} \mathrm{dt}\)
\(\mathrm{I}_{8}+\mathrm{I}_{6}=\left[\frac{\mathrm{t}^{7}}{7}\right]_{0}^{1}=\frac{1}{7}\)
