Let for f ( x ) = 7 tan 8 x + 7 tan 6 x – 3 tan 4 x – 3 tan 2 x , 1 = 0 / 4 f ( x ) d x and 2 = 0 4 x f ( x ) d x . Then 7 I 2 + 12 I 2 is equal to : [2025]

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Solution

Q.
Let for $f (x) = 7 \tan^{8} x + 7 \tan^{6} x - 3 \tan^{4} x - 3 \tan^{2} x$ , $I_{1} = \int_{0}^{π / 4} f (x) d x$ and $I_{2} = \int_{0}^{π / 4} x f (x) d x$ . Then $7 I_{1} + 12 I_{2}$ is equal to : [2025]

1 2

2 $π$

3 1

4 $2 π$

Ans.
(3)

We have, $f (x) = 7 \tan^{8} x + 7 \tan^{6} x - 3 \tan^{4} x - 3 \tan^{2} x$

$= 7 \tan^{6} x (\tan^{2} x + 1) - 3 \tan^{2} x (\tan^{2} x + 1)$

$= (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x$

$I_{1} = \int_{0}^{π / 4} (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x d x$

Putting tan x = t $\Rightarrow \sec^{2} x d x = d t$

$I_{1} = \int_{0}^{1} (7 t^{6} - 3 t^{2}) d t = {[t^{7} - t^{3}]}_{0}^{1} = 0$

$I_{2} = \int_{0}^{π / 4} x (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x d x$

$= \int_{0}^{1} (7 t^{6} - 3 t^{2}) \tan^{- 1} t d t$

$= {[\tan^{- 1} t (t^{7} - t^{3})]}_{0}^{1} - \int_{0}^{1} \frac{t^{7} - t^{3}}{1 + t^{2}} d t = 0 - \int_{0}^{1} \frac{t^{3} (t^{4} - 1)}{1 + t^{2}} d t$

$= - \int_{0}^{1} t^{3} (t^{2} - 1) d t = - \int_{0}^{1} (t^{5} - t^{3}) d t = - {[\frac{t^{6}}{6} - \frac{t^{4}}{4}]}_{0}^{1} = \frac{1}{12}$

$∴ 7 I_{1} + 12 I_{2} = 0 + 12 (\frac{1}{12}) = 1$ .

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