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NIMCET 2013 Question Paper with solutions

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Question : 6 of 120

Marks: +1, -0

If

I_n

\int\limits_{0}^{\pi/4} \tan^n

I_8 + I_6

equals :

1/4
1/5
1/6
1/7

Solution:

Given

I_n = \int\limits^{\pi/4} \tan^n \theta \, d\theta

\Rightarrow I_n = \int\limits_{0}^{\pi/4} \tan^{n-2}\theta \tan^2\theta \, d\theta

\Rightarrow I_n = \int\limits_{0}^{\pi/4} \tan^{n-2}\theta (\sec^2\theta - 1) \, d\theta

[∵

\tan^2\theta = (\sec^2\theta - 1)

]

\Rightarrow I_n = \int\limits_{0}^{\pi/4} \tan^{n-2}\theta \sec^2\theta - \int\limits_{0}^{\pi/4} \tan^{n-2}\theta \, d\theta

\Rightarrow I_n + I_{n-2} = \int\limits_{0}^{1} t^{n-2} \, dt = \left[ \frac{t^{n-1}}{n-1} \right]_{0}^{1}

[∵ let

\tan\theta = t]

\Rightarrow I_n + I_{n-2} = \frac{1}{n-1}

Now, putting n = 8,we get

I_8 + I_{8-2} = \frac{1}{8-1}

\Rightarrow I_8 + I_6 = \frac{1}{7}

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